Type – 01

Short Answer Type Questions :

1 ) Which of the following polynomial is a quadratic polynomial .

( i ) x2 – 7x + 2

( ii ) 7x5 – x ( x + 2 )

( iii ) 2x ( x + 5 ) + 1

( iv ) 2x – 1

Solution : ( i ) x2 – 7x + 2 and ( iii ) 2x ( x + 5 ) + 1

Explanation : These polynomial have maximum power of varible ( x ) is 2. That is why these polynomials are quadratic polynomials.

2 ) Which of the following equations can be written in the form of ax2 + bx + c = 0 , where a, b, and c are real numbers and a is not equal to zero ?

( i ) x – 1 + 1 / x = 6

( ii ) x+ 3 / x = x 2

( iii ) x2 – 6√ x + 2 = 0

( iv ) ( x – 2 )2 = x2 – 4x + 4

Answer : ( i ) x – 1 + 1 / x = 6

Explanation : ( i ) x – 1 + 1 / x = 6

( x2 – x + 1 ) / x = 6

( x2 – x + 1 ) = 6x

x2 – x + 1 – 6x = 0

x2 – 7x + 1 = 0

Hence, this equation can be expressed in the form of ax2 + bx + c = 0

( ii ) x+ 3 / x = x 2

x3 + 3 = x2

x3 + 3 – x2 = 0

This expression has maximum power of the variable ( x ) is 3 which does satisfy the condition of the quadratic equation, ax2 + bx + c = 0

( iii ) x2 – 6 x + 2 = 0 , it contains  x. That’s why it can not be expressed in the form of ax2 + bx + c = 0

( iv ) ( x – 2 )2 = x2 – 4x + 4. It is not an equation as it ( x – 2 )2 is expressed according to the identity of ( a – b )2 = a2 – 2ab – b2

3 ) Determine the power of the variable in which the equation x6 – x3 – 2 = 0 will become a quadratic equation.

Solution : x6 – x3 – 2 = 0

let x3 = a, then

a2 – a – 2 = 0

Hence power is 3.

4 ) Justify and write whether 1 and -1 are the roots of x2 + x + 1 = 0.

Solution :

For justification :

The given equation is x2 + x + 1 = 0.

LHS = x2 + x + 1

Putting x = 1, we get

LHS = ( 1 )2 + 1 + 1 = 1 + 2 = 3 RHS

Again, put x = -1,

LHS = ( -1 )2 + ( -1 ) + 1 = 1 -1 + 1 = 1 RHS

Hence, 1 and -1 are NOT the roots of the given equation.

5 ) Justify and write whether 5 / 6 and 4 / 3 are the roots of the equation, x + 1/ x = 13 / 6 .

Solution :

For Justification :

x + 1/ x = 13 / 6 .

LHS = x + 1/ x

Put x = 5 / 6 , we get

LHS = 5 / 6 + 6 / 5 = ( 25 + 36 ) / 30 = 61 / 30 RHS

Put x = 4 / 3, we get

LHS = 4 / 3 + 3 / 4 = ( 16 + 9 ) / 12 = 25 / 12 RHS

Hence, 5 / 6 and 4/ 3 are NOT the roots of the given equation.

6 ) Justify and write whether – √ 3 and 2√ 3 are the roots of the quadratic equation x23 x – 6 = 0.

Solution :

For justification :

x23 x – 6 = 0.

LHS = x23 x – 6

Put x = – √ 3 , we get

LHS = ( – √ 3 )2 – √ 3 ( – √ 3 ) – 6 = 3 + 3 – 6 = 6 – 6 = 0 = RHS

Again, put x = 2√ 3, we get

LHS = ( 2√ 3 )2√ 3( 2√ 3 ) – 6 = ( 4 * 3 ) – ( 2 * 3 ) – 6 = 12 – 6 – 6 = 12 – 12 = 0 = RHS

Hence, – √ 3 and 2√ 3 are the roots of the given quadratic equation.

7 ) Find the value of k for which 2 / 3 will be a root of the quadratic equation7x2 + kx – 3 = 0.

Solution : 7x2 + kx – 3 = 0.

Since, 2 / 3 is a root of the above equation, therefore, we put x = 2 / 3, we get

7 ( 2 / 3 )2 + k ( 2 / 3 ) – 3 = 0

7 ( 4 / 9 ) + k ( 2 / 3 ) – 3 = 0

28 / 9 + 2k / 3 – 3 = 0

2k / 3 = 3 – 28 / 9

2k / 3 = ( 27 – 28 ) / 9

2k / 3 = 1 / 9

k = ( 1 / 9 ) * ( 3 / 2 )

k = 1 / 6

8 ) If the quadratic equaiton 3x2 + 11x – 4 = 0 has real roots, then determine then with the help of Sridhar Acharya Formula.

Solution : The given quadratic equaiton is 3x2 + 11x – 4 = 0

By comparing above equation with the general form of the quadratic eqaution ax2 + bx + c = 0 , we get

a = 3 , b = 11, c = – 4

Discriminant ( D ) = ( b2 – 4 ac ) = {112 – 4 ( 3 ) ( – 4 ) = ( 121 + 48 ) = 169

The value of discriminant is greater than zero, hence the roots are real

From Sridhar Acharya’s Formula, we have

x = { – b ±  ( b2 – 4 ac ) } / 2a

x = [ -11 ± 169 ] / ( 2 ) ( 3 )

x = [ -11 ± 13 ] / 6

Either, x = ( -11 + 13 ) / 6

x = 2 / 6

x = 1 / 3

Or

x = ( -11 – 13 ) / 6

x = – 24 / 6

x = – 4

The required solution is 1 / 3 or – 4.

9 ) If the quadratic equation ( x – 2 )( x + 4 ) + 9 = 0 has real roots, then determine then with the help of Sridhar Acharya Formula.

Solution : The given equation is ( x – 2 )( x + 4 ) + 9 = 0

or, x ( x + 4 ) – 2 ( x + 4 )+ 9 = 0

or, x2 + 4x – 2x – 8+ 9 = 0

or, x2 + 2x + 1 = 0

By comparing above equation with the genreal form of quadratic equaiton ax2 + bx + c = 0 , we get

a = 1 , b = 2 , c = 1

Discriminant ( D ) =  ( b2 – 4 ac ) = { 22 – 4 (1)(1)} =  ( 4 – 4 ) = 0

Since, the vlaue of D is equal to zero, therefore the roots are real and equal.

From Sridhar Acharya’s Formula, we have

x = { – b ±  ( b2 – 4 ac ) } / 2a ]

or, x = { – 2 ± 0 } / 2 ( 1 )

or, x = ( – 2 + 0 ) / 2

or, x = -2 / 2

or, x = – 1

The required solution is x = -1 , -1

10 ) If the quadratic equaiton ( 4x – 3 )2 – 2( x + 3 ) = 0 has real roots, then determine then with the help of Sridhar Acharya Formula.

Solution : The given equation is

( 4x – 3 )2 – 2( x + 3 ) = 0

or, ( 4x )2 – 2 ( 4x ) ( 3 ) + 3 2 – 2x – 6 = 0

or , 16x2 – 24x + 9 – 2x – 6 = 0

or, 16x2 – 26x + 3 = 0

By comparing above equation with general form of the quadratic equation ax2 + bx + c = 0 , we get

a = 16, b = -26 , c = 3

Discriminant ( D ) = ( b2 – 4 ac ) = { ( – 26 )2 -4 (16 ) ( 3 ) } = ( 676 – 192 ) = 484

The value of D is positive, therefore the roots are real and unequal.

From Sridhar Acharya’s Formula, we have

x = { – b ±  ( b2 – 4 ac ) } / 2a

or, x = [ – ( – 26 ) ±  484 ] / 2 ( 16 )

or, x = ( 26 ± 22 ) / 32

Either x = ( 26 + 22 ) / 32

x = 48 / 32 = 3 / 2

Or x = ( 26 – 22 ) / 32 = 4 / 32 = 1 / 8

The required solutions are 3/ 2 and 1 / 8

11 ) If the quadratic equaiton 3x2 + 2x + 1 = 0 has real roots, then determine them with the help of Sridhar Acharya Formula.

solution : The given equaiton is 3x2 + 2x + 1 = 0

By comparing above equation with the general form of the quadratic equation ax2 + bx + c = 0, we get

a = 3, b = 2, c = 1

Discriminant ( D ) = ( b2 – 4 ac ) = 22 – 4 ( 3 ) ( 1 ) = 4 – 12 = – 8

Since, the value of D is negative, the roots are not real.

12 ) If the quadratic equaiton 10x2 – x – 3 = 0 has real roots, then determine then with the help of Sridhar Acharya Formula.

Solution : The given equation is 10x2 – x – 3 = 0

By comparing above equation with the general form of quadratic equation ax2 + bx + c = 0, we get

a = 10, b = -1, c = -3

Discriminant ( D ) = ( b2 – 4 ac ) = ( -1 )2 – 4 ( 10 ) ( -3 ) = 1 + 120 = 121

From Sridhar Acharya’s Formula, we have

x = { – b ±  ( b2 – 4 ac ) } / 2a

or, x = { – ( -1 ) ±  121 } / 2 ( 10 )

or, x = ( 1 ± 11 ) / 20

Either, x = ( 1 + 11 ) / 20 = 12 / 20 = 3 / 5

Or x = ( 1 – 11 ) = – 10 / 20 = – 1 / 2

13 ) If the quadratic equaiton 10x2 – x + 3 = 0 has real roots, then determine then with the help of Sridhar Acharya Formula.

Solution : The given quadratic equation is 10x2 – x + 3 = 0

By comparing above equation with the general form of the quadratic equation ax2 + bx + c = 0, we get

a = 10, b = – 1 , c = 3

Discriminant ( D ) = ( b2 – 4 ac ) = ( -1 )2 – 4( 10 ) ( 3 ) = 1 – 120 = – 119

Since, the value of D is negative, the roots are not real .

14 ) If the quadratic equaiton 25x2 – 30x + 7 = 0 has real roots, then determine them with the help of Sridhar Acharya Formula.

Solution : The given equation is 25x2 – 30x + 7 = 0

By comparing above equation with the general form of the quadratic equation ax2 + bx + c = 0 , we get

a = 25, b = – 30, c = 7

Discriminant ( D ) = ( b2 – 4 ac ) = ( – 30 )2 – 4 ( 25 ) ( 7 ) = 900 – 700 = 200

From Sridhar Acharya’s Formula, we have

x = { – b ±  ( b2 – 4 ac ) } / 2a

or, x = {- ( – 30 ) ±  200 } / 2 (25 ) = ( 30 ± 10 2 ) / 2 ( 25 ) = 10 ( 3 ±  2 ) / 50 = ( 3 ±  2 ) / 5

Either x = ( 3 +  2 ) / 5

or x = ( 3 –  2 ) / 5

Hence, the required roots are ( 3 +  2 ) / 5 and ( 3 –  2 ) / 5

15 ) If the quadratic equaiton ( 4x – 2 )2 + 6x = 25 has real roots, then determine then with the help of Sridhar Acharya Formula.

Solution : The given equation is given by

( 4x – 2 )2 + 6x = 25

or, ( 4x )2 – 2 ( 4x ) ( 2) + ( 2 )2 + 6x = 25

or, 16x2 – 16x + 4 + 6x = 25

or, 16x2 – 10x + 4 – 25 = 0

or, 16x2 – 10x – 21 = 0

By comparing above equation with the general form of the quadratic equation ax2 + bx + c = 0 , we get

a = 16, b = -10, c = – 21

Discriminant ( D ) = ( b2 – 4 ac ) = ( – 10 )2 – 4 ( 16 )( – 21 ) = 100 + 1344 = 1444

From Sridhar Acharya’s Formula, we have

x = { – b ±  ( b2 – 4 ac ) } / 2a

or, x = { – ( -10 ) ± 1444 } / 2 ( 16 )

or, x = ( 10 ± 38 ) / 32

Either x = ( 10 + 38 ) / 16 = 48 / 32 = 3 / 2

or x = ( 10 – 38)/ 32 = – 28 / 32 = – 7 / 8

The required roots are 3 / 2 and – 7 / 8

16 ) Find the value of a, if one root of the quadratic equation x2 + ax + 3 = 0 is 1

Solution: The given quadratic equation is …………………

x2 + ax + 3 = 0

Since 1 is one of the roots of the above equation, we put x = 1

( 1 )2 + a ( 1 ) + 3 = 0

or 1 + a + 3 = 0

or a + 4 = 0

or a = -4

Hence, the required value of a is – 4

17 ) Calculate the value of the other root if one root of the equation x2 – ( 2 + b ) x + 6 = 0 is 2.

Solution : The given quadratic equation is …………………….

x2 – ( 2 + b )x + 6 = 0

By comparing above equation with the general form of the quadratic equation, we get

a = 1, b = – ( 2 + b ), c = 6

Product of the roots = c / a = 6 / 1

or, one roots * other root = 6

or, 2 * other root = 6

or, other root = 6 / 2

or, other root = 3

18 ) What is the value of the other root if one root of the equation 2x2 + kx + 4 = 0 is 2 ?

Solution : The given quadratic equation is …………………….

2x2 + kx + 4 = 0

By comparing above equation with the general form of the quadratic equation, we get

a = 2 , b = k, c = 4

Product of the roots = c / a

one root * other root = 4 / 2

or, 2 * other root = 2

or, other root = 2 / 2

or, other root = 1

19 ) Write the equation if the difference of a proper fraction and its reciprocal is 9 / 20.

Solution : Let the proper fraction be x, then its reciprocal be 1 / x

The required equation is given by

x – 1 / x = 9 / 20

( x2 – 1 ) / x = 9 / 20

20x2 – 20 = 9x

20x2 – 9x – 20 = 0

20 )Write the values of ‘ a ‘ and ‘b’, if the two roots of the equation ax2 + bx + 35 = 0 are -5 and -7.

Solution : The given equation is ……………

ax2 + bx + 35 = 0

Since – 5 and – 7 are the two roots of the above equation, we put the x = – 5 and x = – 7 successively,

a ( -5 )2 + b ( -5 ) + 35 = 0

or, 25a – 5b + 35 = 0 ……………………………………………… ( 1 )

and a ( – 7 )2 + b ( – 7 ) + 35 = 0

or, 49a – 7b + 35 = 0 ……………………………………………… ( 2 )

Multiplying equation ( 1 ) by 7 and equation ( 2 ) by 5, we get

175a – 35b+ 245 = 0 ……………………… ( 3 )

245a – 35b + 175 = 0 …………………. ( 4 )

Now, subtracting equation (4 ) by equation ( 3 ), we get

245a – 175a + 175 – 245 = 0

70 a – 70 = 0

or, 70a= 70

or, a = 70 / 70

or, a = 1

Now putting a = 1 in equation ( 1 ) we get

25a – 5b + 35= 0

or, 25 ( 1 ) – 5b + 35 = 0

or, 25 – 5b + 35 = 0

or, – 5b + 60 = 0

or, – 5b = – 60

or, b = 60 / 5

or, b = 12

Hence, the values of a and b are 1 and 12 respectively.

21 ) Write whether the Sridhar Acharya’s Formula is applicable to solve the equaiton 4x2 + ( 2x + 1 ) ( 2x – 1 ) = 4x ( 2x – 1 ).

Solution : The given equaiton is ………………..

4x2 + ( 2x + 1 ) ( 2x – 1 ) = 4x ( 2x – 1 ).

or, 4x2 + ( 2x )2 – ( 1 )2 = 8x2 – 4x

or, 4x2 + 4x2 – 1 = 8x2 – 4x

or, 8x2 – 8x2 + 4x + 1 = 0

or, 4x + 1 = 0

Since, this is a linear equation and Sridhar Acharya formula is applicable only for the quadratic equation. Hence, Sridhar Acharya formula is not applicable to solve the given equation.

22 ) By applying Sridhar Acharya’s Formula in the equaiton 5x2 + 2x – 7 = 0, it is found that x = ( k ± 12 )/ 10, what will be the value of k ?

Solution : The given equation is ……………………..

5x2 + 2x – 7 = 0

By comparing above equation with the general form of the quadratic equation ax2 + bx + c = 0, we get

a = 5, b = 2, c =-7

Applying Sridhar Acharya formula, we get

x = { – b ±  ( b2 – 4 ac ) } / 2a

or, x = [-2 ±  { ( -2 )2 – 4 ( 5 )(-7 )} ] / 2 ( 5 )

or ,( k ± 12 )/ 10 = [ – 2 ±  ( 4 + 140) ] / 10

or ( k ± 12 )/ 10 = { -2 ±  144 } / 10

or, ( k ± 12 ) / 10 =(-2 ± 12 ) / 10

or, k ± 12 = -2 ± 12

or, k = -2

Hence, the required value of k is – 2.

23 ) Find out the nature of the two roots of the quadratic equation 3x2 + 2√6 x + 2 = 0

Solution : The given equatin is …………………..

3x2 + 26 x+ 2 = 0,

By comparing above equation with the general form of the quadratic equation ax2 + bx + c = 0, we get

a = 3, b = 26, c = 2

Discriminant = b2 – 4ac = ( 26 )2 – 4 ( 3 ) ( 2 ) = 24 – 24 = 0

Hence, the roots are real and equal.

24 ) Find out the nature of the two roots of the quadratic equation 2x2 – 7 x + 9 = 0

Solution : The given equation is ……………….

2x2 – 7x + 9 = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0 , we get

a = 2, b = -7, c = 9

Dicriminant =b2 – 4ac = ( -7 )2 – 4 ( 2 ) ( 9 )= 49 – 72 = – 23

Since, the value of dicriminant is negative, the roots are not real.

25 ) Find out the nature of the two roots of the quadratic equation (2 / 5)x2 – ( 2 / 3 )x + 1 = 0

Solution :

Solution : The given equation is ……………….

( 2 / 5)x2 – ( 2 / 3 )x + 1 = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0 , we get

a = 2 / 5, b = – 2 / 3, c = 1

Dicriminant =b2 – 4ac = ( – 2 / 3 )2 – 4 ( 2 / 5 ) ( 1 )= 4 / 9 – 8 / 5 = ( 20 – 72 ) / 45 = – 52 / 45

Since, the value of discriminant is negative, the roots are not real.

26 ) Calulate the value(s) of k for which the quadratic equation 49x2 + kx + 1 = 0 has real and equal roots.

Solution : The given equation is ……………….

49x2 + kx + 1 = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0 , we get

a = 49, b = k , c = 1

For equal roots, Discriminat = 0

b2 – 4ac = 0

or, k2 – 4 ( 49 ) ( 1 ) = 0

or, k2 = 49 * 4

or, k = ( 49 * 4 )

or, k = ± 7 * 2

or, k = ± 14

27 ) Calulate the value(s) of k for which the quadratic equation 3x2 – 5x + 2k = 0 has real and equal roots.

Solution : The given equation is ……………….

3x2 -5x + 2k = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0 , we get

a = 3, b = -5, c = 2k

For equal roots, Discriminat = 0

b2 – 4ac = 0

or, ( -5 )2 – 4 ( 3 ) ( 2k ) = 0

or, 25 – 24 k = 0

or, – 24k = -25

or, k = 25 / 24

28 ) Calulate the value(s) of k for which the quadratic equation 2x2 + 3x + k = 0 has real and equal roots.

Solution : The given equation is ……………….

2x2 + 3x + k = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0 , we get

a = 2, b = 3 , c = k

For equal roots, Discriminat = 0

b2 – 4ac = 0

or, ( 3 )2 – 4 ( 2 ) ( k ) = 0

or, 9 – 8k = 0

or, – 8k = -9

or, k = 9 / 8

29 ) Calulate the value(s) of k for which the quadratic equation x2 – 2 ( 5 + 2k )x + 3( 7 + 10k ) = 0 has real and equal roots.

Solution : The given equation is ……………….

x2 – 2 ( 5 + 2k )x + 3 ( 7 + 10 k ) = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0 , we get

a = 1, b = – 2 ( 5 + 2k ), c = 3 ( 7 + 10k )

For equal roots, Discriminat = 0

b2 – 4ac = 0

or, { – 2 (5 + 2k ) } 2 – 4 ( 1 ) { 3 ( 7 + 10k ) = 0

or, 4 ( 5 + 2k )2 – 12 ( 7 + 10 k ) = 0

or, 4 { ( 5 )2 + 2 ( 5 ) ( 2k ) + ( 2k )2 } – 84 – 120k = 0

or, 4 ( 25 + 20 k + 4k2 ) – 84 – 120k = 0

or, 100 + 80k + 16k2 – 84 – 120k = 0

or, 16k2 – 40k + 16 = 0

or, 8 ( 2k2 – 5k + 2 ) = 0

or, 2k2 – 5k + 2 = 0

or, 2k2 – ( 4 + 1 ) k + 2 = 0

or, 2k2 -4k – k + 2 = 0

or, 2k ( k – 2 ) – 1 ( k – 2 ) = 0

or, ( k – 2 ) ( 2k – 1 ) = 0

Either, k – 2 = 0

k = 2

or, 2k – 1 = 0

or , 2k = 1

or, k = 1 / 2

The required values of k are 2 and 1 / 2

30 ) Calulate the value(s) of k for which the quadratic equation ( 3k + 1 )x2 + 2( k + 1 )x + k = 0 has real and equal roots.

Solution : The given equation is ……………….

( 3k + 1 )x2 + 2( k + 1 )x + k = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0 , we get

a = ( 3k + 1 ), b = 2( k + 1 ), c = k

For equal roots, Discriminat = 0

b 2 – 4ac = 0

or , 22 ( k + 1 )2 – 4 ( 3k + 1 )( k ) = 0

or, 4 ( k2 + 2k + 1 ) – 12k2 – 4k = 0

or , 4k2 + 8k + 4 – 12k2 – 4k = 0

or, – 8k2 + 4k + 4 = 0

or, -4 ( 2k2 – k – 1 ) = 0

or, 2k2 – k – 1 = 0

or, 2k2 – 2k + k – 1 = 0

or, 2k( k – 1 ) + 1 ( k – 1 ) = 0

or, ( k – 1 ) ( 2k + 1 ) = 0

Either, k – 1 = 0

or k = 1

Or, ( 2k + 1 ) = 0

or, 2k = -1

or, k = – 1 / 2

The required values of k are 1 and – 1 / 2

31 ) Form the quadratic equation from two roots 4, 2.

Solution : Sum of the roots = 4 + 2 = 6

Product of the roots = 4 * 2 = 8

The required equation is given by ………………..

x2 – ( sum of the roots )x + ( product of the roots ) = 0

Or, x2 – 6x + 8 = 0

32 ) Form the quadratic equation from two roots – 4, – 3

Solution :

Sum of the roots = ( – 4 ) + ( -3 ) = -7

Product of the roots = ( -4 ) * ( -3 ) = 12

The required quadratic equation is given by…………………………..

x2 – ( sum of the roots )x + ( product of the roots ) = 0

x2 – ( -7 )x + 12 = 0

x2 + 7x + 12 = 0

33 ) Form the quadratic equation from two roots – 4 , 3

Solution :

Sum of the roots = ( – 4 ) + 3 = – 4 + 3 = – 1

Product of the roots = ( – 4 )* ( 3 ) = – 12

The required equation is given by……………………………….

x2 – ( sum of the roots )x + ( product of the roots ) = 0

x2 – ( – 1 )x + ( – 12 ) = 0

x2 + x – 12 = 0

34 ) Form the quadratic equation from two roots 5, -3.

Solution :

Sum of the roots = 5 + ( -3 ) = 5 – 3 = 2

Product of the roots = 5 * ( – 3 ) = – 15

The required equation is given by …………………………….

x2 – ( sum of the roots )x + ( product of the roots ) = 0

x2 – 2x + ( – 15 ) = 0

x2 – 2x – 15 = 0

35 ) Find m for which the two roots of the quadratic equaiton 4x2 + 4 ( 3m – 1 ) x + ( m + 7 ) = 0 are reciprocal to each other.

Solution : Let one root is ‘ p ‘ then other roots is 1 / p

The given equation is ………………………..

4x2 + 4 ( 3m – 1 )x + ( m + 7 ) = 0

By comparing above equation with the general form of the quadratic equaiton ax2 + bx + c = 0, we get

a = 4, b = 4 ( 3m – 1 ) = 12m – 4, c = ( m + 7 )

According to problem,

Product of the roots = p * ( 1 / p )

or, c / a = 1

or, c = a

or, m + 7 = 4

or, m = 4 – 7

or, m = – 3

36 ) If two roots of the quadratic equation ( a2 + b2 ) x2 – 2c ( ac + bd )x + ( c 2 + d 2 ) = 0 are equal then prove that a / b = c / d

Solution : The given equation is ……………….

( a2 + b2 ) x2 – 2 (ac +bd ) x + ( c2 + d2 ) = 0

By comparing above equation with the general form of the quadratic equaiton Ax2 + Bx + Q = 0, we get

A = ( a2 + b2 ) , B = -2 ( ac + bd ) and Q = ( c2 + d2 )

Since, the roots are equal, therefore,

B2 – 4AQ = 0

or, { -2 ( ac + bd ) }2 – 4 ( a2 + b2 )( c2 + d2 ) = 0

or, { -2( ac + bd ) }2 = 4 ( a2 + b2 )( c2 + d2 )

or, 4(ac + bd )2 = 4 ( a2 + b2 )( c2 + d2 )

or, ( a2 c2 + b2 d2 + 2abcd ) = ( a2 c2 + a2d2 + b2c2 + b2d2 )

or, 2abcd = a2d2 + b2c2

or, ( ad )2 + ( bc ) 2 – 2( ad ) ( bc ) = 0

or, ( ad – bc )2 = 0

or, (ad – bc ) = 0

or, ad = bc

or, a / b = c / d

( Proved )

37 ) If the two roots of the quadratic equation ( b – c ) x2 + ( c – a )x + ( a – b ) = 0 are equal, then prove that 2b = a + c

Solution : The given equation is ……………….

( b – c ) x2 + ( c – a ) x + ( a – b ) = 0

By comparing above equation with the general form of the quadratic equaiton Ax2 + Bx + Q = 0, we get

A = ( b – c ) , B = ( c – a ) and Q = ( a – b )

Since, the roots are equal, therefore,

B2 – 4AQ = 0

( c – a )2 – 4 ( b – c )( a – b ) =0

or, c2 – 2ac + a2 – 4( ab – b2 -ac + bc ) = 0

or, c2 – 2ac + a2 – 4ab + 4b2 + 4ac – 4bc = 0

or, a2 + c2 + 4b2 + 2ac – 4ab – 4bc = 0

or, ( a )2 + ( c ) 2 + ( – 2b ) 2 + 2 ( a ) ( c ) + 2 ( c ) ( – 2b ) + 2 ( – 2b ) ( a ) = 0

Or, ( a + c – 2b ) 2 = 0

or, ( a + c – 2b ) = 0

or, a + c = 2b

( Proved )

38 ) Prove that, the quadratic equation 2 ( a2 + b2 ) x2 + 2 ( a + b )x + 1 = 0 has no real roots if a ≠ b.

Solution : For no real roots, the discriminaant should be negative or less than zero.

Therefore,

{ 2 ( a + b ) }2 – 4{ 2 ( a2 + b2 )}( 1 ) < 0

or, 4 ( a 2 + b2 + 2ab ) – 8 ( a + b2 ) < 0

or, a2 + b2 + 2ab – 2 ( a2 + b2 ) < 0

or, a2 + b2 + 2ab – 2a2 – 2b2 < 0

or, -a2 – b2 + 2ab < 0

or, – ( a2 + b2 – 2ab )< 0

or, ( a2 + b2 – 2ab ) > 0

or, ( a – b )2 > 0

or, ( a – b ) > 0

or, a > b

Hence, a ≠ b ( Proved )

39 ) If the two roots of the quadratic equation 5x2 + 2x -3 = 0 are α and β then determine the value of α2 + β2.

Solution :

The given equation is 5x2 + 2x – 3 = 0

By comparing above equation with the general form of of the quadratic equation, we get

a = 5, b = 2 and c = -3

Sum of the roots = – b / a

or, ( α + β ) = – 2 / 5 ——————————————————-( 1 )

Product of the Roots = c / a

or, αβ = – 3 / 5 ——————————————————( 2 )

Now ( α2 + β2 ) = ( α + β )2 – 2αβ

= ( – 2 / 5 )2 – 2 ( – 3 / 5 )

= ( 4 / 25 ) + ( 6 / 5 )

= ( 4 + 30 ) / 25

= 34 / 25

40 ) If the two roots of the quadratic equation 5x2 + 2x -3 = 0 are α and β then determine the value of α3 + β3.

Solution :

The given equation is 5x2 + 2x – 3 = 0

By comparing above equation with the general form of of the quadratic equation, we get

a = 5, b = 2 and c = -3

Sum of the roots = – b / a

or, ( α + β ) = – 2 / 5 ——————————————————-( 1 )

Product of the Roots = c / a

or, αβ = – 3 / 5 ——————————————————( 2 )

Now ( α3 + β3 ) = ( α + β )3 – 3αβ ( α + β )

= ( – 2 / 5 )3 – 3 ( – 3 / 5 ) ( – 2 / 5 )

= – ( 8 / 125 ) – ( 18 / 25 )

= ( – 8 – 90 ) / 125

= – 98 / 125

41 ) If the two roots of the quadratic equation 5x2 + 2x -3 = 0 are α and β then determine the value of 1 /α + 1 / β.

Solution :

The given equation is 5x2 + 2x – 3 = 0

By comparing above equation with the general form of of the quadratic equation, we get

a = 5, b = 2 and c = -3

Sum of the roots = – b / a

or, ( α + β ) = – 2 / 5 ——————————————————-( 1 )

Product of the Roots = c / a

or, αβ = – 3 / 5 ——————————————————( 2 )

1 / α + 1 / β = ( α2 + β2 ) / αβ = { ( α + β )2 – 3 αβ } / αβ

= { ( – 2 / 5 ) 2 – ( – 3 / 5 ) } / ( – 3 / 5 ) }

= { 4 / 25 + 6 / 5} / ( – 3 / 5 )

= { ( 4 + 30 ) / 25 } / ( – 3 / 5 )

= ( 34 / 25 ) / ( – 3 / 5 )

= – ( 34 * 3 ) / ( 25 * 5 )

= – 102 / 125

42 ) If the two roots of the quadratic equation 5x2 + 2x -3 = 0 are α and β then determine the value of α2 / β + β2 / α.

Solution :

The given equation is 5x2 + 2x – 3 = 0

By comparing above equation with the general form of of the quadratic equation, we get

a = 5, b = 2 and c = -3

Sum of the roots = – b / a

or, ( α + β ) = – 2 / 5 ——————————————————-( 1 )

Product of the Roots = c / a

or, αβ = – 3 / 5 ——————————————————( 2 )

Now

α2 / β + β2 / α. = ( α3 + β3 ) / αβ

= { ( α + β )3 – 3αβ ( α + β ) } / αβ

= {( – 2 / 5 )3 – 3 ( – 3 / 5 ) ( – 2 / 5 )} / ( – 3 / 5 )

= { – ( 8 / 125 ) – ( 18 / 25 ) } / ( – 3 / 5 )

= { ( – 8 – 90 ) / 125 } / ( – 3 / 5 )

= ( – 98 / 125 ) / ( – 3 / 5 )

= ( 98 * 3 ) / ( 125 * 5 )

= 294 / 625

43 ) If one root of hte equaiton ax2 + bx + c = 0 is twice of the other, then show that , 2b2 = 9ac.

Solution : Let one root is p and other root is 2p

Sum of the roots = – b / a

or, p + 2p = – b / a

or, 3p = – b / a

or, p = – b / 3a …………………………………………… ( 1 )

and the product of the roots = c / a

or, p * 2p = c / a

or, 2p2 = c / a

or, 2 ( – b / 3a )2 = c / a

or, 2 ( b2 / 9 a2 ) = c / a

or, 2b2 = 9ac ( Proved )

44 ) Form the quadratic equation whose roots are rciprocals to the roots of the equation x2 + px + 1 = 0

Solution : The given equaiton is x2 + px + 1 = 0

Let ‘q’ and ‘r’ are the roots of the above equation, then

q + r = p / 1 = -p

and qr = 1

The roots of the required quadratic equation are 1 / q and 1 / r.

sum of the roots = 1 / q + 1 / r = ( q +r ) / qr = -p / 1 = – p

product of the roots = ( 1 / q ) ( 1 / r ) = / ( qr ) = 1 / 1 = 1

The required quadratic equation is given by

x2 – ( sum of the roots )x + ( product of the roots ) = 0

x2 – ( – p ) x + 1 = 0

x2 + px + 1 = 0

45 ) Determine the quadratic equation whose roots are square of the roots of the equation x2 + x + 1 = 0

solution : The given equation is x2 + x + 1 = 0

Let p and q are the roots of the given equation , then

sum of the roots = p + q = – 1

and products of the roots = pq = 1

The roots of the required quadratic equation are p2 and q2

Now, the sum of the roots = p2 + q2 = ( p + q )2 – 2pq = ( -1 )2 – 2 ( 1 ) = 1 – 2 = -1

product of the roots = p2q2 = ( pq )2 = 12 = 1

The required quadratic equation is given by

x2 – ( sum o fthe roots )x + ( product o fthe roots ) = 0

x2 – ( – 1 ) x + 1 = 0

x2 + x + 1 = 0

46 ) Write the quadratic equation if the sum of the roots is 14 and the product of them is 24.

Solution : The required quadratic equation is given by

x2 – ( sum of the roots )x + ( product of the rots ) = 0

x2 – 14x + 24 = 0

47 ) If the sum and the product of the two roots of the equation kx2 + 2x + 3k = 0 ( k ≠ 0 ) are equal, write the value of k.

Solution : The given quadratic equation is kx2 + 2x + 3k = 0

sum of the roots = product of the roots

2 / k = 3 k / k

or, 2 / k = 3

or, k = 2 / 3

The required value of is 2 / 3.

48 ) If the two roots of the equation x2 – 22x + 105 = 0 are α and β then write the value of ( α – β ).

Solution : The given equation is x2 – 22x + 105 = 0

( α + β ) = – ( -22 / 1 ) = 22

α * β= 105 / 1 = 105

( α – β )2 = ( α + β )2 – 4 α * β

or, ( α – β )2 = ( 22 )2 – 4 ( 105 )

or, ( α – β )2 = 484 – 420

or, ( α – β )2 = 64

or, ( α – β ) = ± 8

49 ) If the sum of the two roots of the equation x2 – x = k ( 2x – 1 ) is zero, write the value of k.

Solution : The given equation is

x2 – x = k ( 2x – 1 )

or, x2 – x = 2kx – k

or, x2 -x – 2kx +k = 0

or, x2 – (1 + 2k )x + k = 0

Sum of the roots = 0

or, – { – ( 1 + 2k ) } / 1 = 0

or, 1 + 2k = 0

or, 2k = – 1

or, k = – 1 / 2

50 ) If one of the roots of the two equations x2 + bx + 12 = 0 and x2 + bx + q = 0 is 2, write the value of q.

Solution : Since 2 is the roots of the equation x2 + bx + 12 = 0 ,

22 + b ( 2 ) + 12 = 0

or, 4 + 2b + 12 = 0

or, 2b + 16 = 0

or, 2b = – 16

or, b = -16 / 2

or, b = – 8

Now, putting x = 2 and b = – 8 in the equation x2 + bx + q = 0 , we get

22 + ( – 8 ) ( 2 ) + q = 0

or, 4 – 16 + q = 0

or, q – 14 = 0

or, q = 14

51 ) Determine the value of a for which the equation ( a – 2 )x2 + 3x + 5 = 0 will not be a quadratic equation.

Solution : For non quadratic equaiton,

Coefficient of x2 = 0

or, ( a – 2 ) = 0

or, a = 2

Type – 02

PROBELM SUMS

1 ) The difference of two postive whole numbers is 3 and the sum of their squares is 117, find the two numbers.

Solution : Let us assume that the greater number among two number is x, then the smaller will be ( x – 3 ).

According to problem,

( x – 3 )2 + x2 = 117

x2 – 2 ( x ) ( 3 ) + 3 2 + x2 = 117

x2 – 6x + 9 + x2 – 117 =0

2x2 – 6x – 108 = 0

2 ( x2 – 3x – 54 ) = 0

x2 – 3x – 54 = 0

x2 – ( 9 – 6 )x – 54 = 0

x2 – 9x + 6x – 54 = 0

x ( x – 9 ) + 6 ( x – 9 ) = 0

( x – 9 ) ( x + 6 ) = 0

Either, ( x – 9 ) = 0

x = 9

or ( x + 6 ) = 0

x = – 6

Since, the numbers are positve

Therefore, x -6

The required numbers are 9 and ( 9 – 3 ) = 6

2 ) The base of a triangle is 18m more than two times of its height, if the area of the triangle is 30 sq.m, then determine the height of it.

Solution :

In the above figure, let h = height of the triangel and b = base of the triangle, then, b = ( 2h + 18 )m

Area = 360 sq.m

( 1/ 2 ) * b * h = 360

bh = 360 * 2

( 2h + 18 )h = 720

2 ( h2 + 9h ) = 720

h2 + 9h = 720 / 2

h2 + 9h = 360

h2 + 9h – 360 = 0

h2 + ( 24 – 15 )h – 360 = 0

h2 + 24h – 15h – 360 = 0

h ( h + 24 ) – 15 ( h + 24 ) = 0

( h + 24 ) ( h – 15 ) = 0

Either, ( h + 24 ) = 0

h = – 24

or

h – 15 = 0

h = 15

Since, height of the triangle can never be in negative, therefore

h -24 m

therefore h = 15 m

Height = 15 m

Base = 2h + 18 = 2 ( 15 ) + 18 = 30 + 18 = 48m

3 ) If five times of a positive whole number is less by 3 than twice of its square, then let us determine the number.

Solution : Let the number be ‘x’.

According to problem,

2x2 – 3 = 5x

or, 2x2 – 5x – 3 = 0

or 2x2 – ( 6 – 1 )x – 3 = 0

or, 2x2 – 6x + x – 3 = 0

or, 2x ( x – 3 ) + 1 ( x – 3 ) = 0

or, ( x – 3 ) ( 2x + 1 ) = 0

Either, x – 3 = 0

or, x = 3

Or, 2x + 1= 0

or, 2x = -1

or, x = – 1 / 2

Since, the required number is positive whole number , x ≠ – 1 / 2

Therefore, x = 3

The required number should be 3

4 ) The distance betwen two places is 200 km; the time taken by a motor car for going from one place to another is less by 2 hrs than the time taken by a jeep car. If the speed of the motor car is 5 km / hr more than the speed of the jeep car, calculate the speed of the motor car.

Solution : Let the speed of the motor car be ‘x’ km / hr.

Speed of the jeep car should be ( x – 5) km /hr.

For motor car :

speed( v ) = x km / hr

distance ( s ) = 200 km

time ( t ) = s / v = ( 200 / x ) hr

For Jeep car :

Speed ( v ) = ( x – 5 ) km/hr

distance ( s) = 200 km

Time ( t ) = s / v = 200 / ( x – 5 ) hr

According to problem,

200 / ( x – 5 ) – 200 / x= 2

or 200 { 1 / x – 1 / ( x – 5 ) } = 2

or, 100 { ( x – ( x – 5 ) } / x ( x – 5 ) = 21

or, 100 { ( x – x+ 5 ) / ( x2 – 5x ) } = 1

or, 100 { 5 / ( x2 – 5x ) = 1

or, x2 – 5x = 500

or, x2 – 5x – 500 = 0

or, x2 – ( 25 – 20 )x – 500 = 0

or, x2 – 25x + 20 x – 500 = 0

or, x ( x – 25 ) + 20 ( x – 25 ) = 0

or, ( x – 25 ) ( x + 20 ) = 0

Either, ( x – 25 ) = 0

or, x = 25

Or, x + 20 = 0

or, x = -2

But speed can never be negative, thefore the value of x should not be -20 km /hr.

Hence, x = 25

The required speed of the motor car is 25 km/hr.

5 ) The area of the Amita ‘s rectangular land is 2000sqm and peimeter of it is 180 m. Find the length and breadth of the Amita’s land.

Solution : Let l = length and b = breadth

Area = 2000 sq ma

or, lb = 2000

or, l = 2000 / b …………………………………….. ( 1 )

and perimeter = 180 m

or, 2 ( l + b ) = 180

or, l + b = 180 / 2

or, 2000 / b + b = 90

or, ( 2000 + b2 ) / b = 90

or, b2 + 2000 = 90b

or, b2 – 90b + 2000 = 0

or, b2 – ( 50 + 40 ) b + 2000 = 0

or, b2 – 50b – 40 b + 2000 = 0

or, b ( b – 5 ) – 40 ( b – 50 ) = 0

or, ( b – 50 ) ( b – 40 ) = 0

Either, ( b – 50 ) = 0

or, b = 50

or, ( b – 40 ) = 0

or, b = 40

If we put b = 40 in equation ( 1 ) , we get,

l = 2000 / b = 2000 / 40 = 50

and we get by putting b = 50 in equation ( 1 ),

l = 2000 / 50 = 40

but lenght is generally greater than breadth.

Hence, required length is 50 m while breadth is 40 m.

6 ) The tens digit of a two digital number is less by 3 than the units digit. If the product of the two digits is subtracted from the number, the result is 15. Find the unit digit of the number.

Solution : Let the unit digit be ‘ x ‘.

tens digit = x – 3

The number should be { 10(x – 3 ) + x } = ( 10 x – 30 + x } = 11x – 30

Product of the digits = x ( x – 3 ) = x2 – 3x

According to problem,

Number – product of the digits =15

( 11x – 30 ) – ( x2 – 3x ) = 15

or, 11x – 30 – x2 + 3x – 15 = 0

or, -x2 + 14x -45 = 0

or, – ( x2 – 14x + 5) = 0

or, x2 – 14x + 45 = 0

or, x2 – ( 9 + 5 ) x + 45 = 0

or, x2 – 9x – 5x + 45 = 0

or, x ( x – 9 ) – 5 ( x – 9 ) = 0

or, ( x – 9 ) ( x – 5 ) = 0

Either, x – 9 = 0

or, x = 9

OR, x – 5 = 0

or, x = 5

The required unit digit is 5 or 9.

7 ) There are two pipes in a water resevoir of oure school. Two pipes together take 11 1 / 9 minutes to fill the reservior. If the two pipes are opened separately, then one pipe would take 5 minutes more time than that of the other pipe. Find out the time taken to fill the reservoir separately by each of the pipes.

Solution : let time taken to fill the reservoir by first pipe is x minutes, then time taken by the seocnd pipe should be ( x + 5 ) minutes.

In x minute, first pipe can fill = 1 complete reservoir

In 1 minute, first pipe can fill = 1/ x part of the reservoir

Similarly, in 1 minute, second pipe can fill = 1 / ( x + 5 ) part of the reservoir.

In 1 minute, both pipes can fill = { 1 / x + 1 / ( x + 5 ) } part of the reservoir.

= { ( x + 5 + x ) / ( x ( x + 5 ) }

= ( 2x + 5 ) / ( x2 + 5 x )

Now,

Time required to fill ( 2x + 5 ) / ( x2 + 5x ) part of the reservior = 1 minute

Time required to fill 1 complete resevoir by both pipe = 1 / { ( 2x + 5 ) / ( x2 + 5x ) minutes. = ( x2 + 5x ) / ( 2x + 5 ) minutes

According to problem,

( x2 + 5x ) / ( 2x + 5 ) = 11 1 / 9

or, ( x2 + 5x ) / ( 2x + 5 ) = 100 / 9

or, 9x2 + 45x = 200x + 500

or, 9x2 + 45x – 200x – 500 = 0

or, 9x2 – 155x – 500 = 0

or, 9x2 – ( 180 – 25 )x – 500 = 0

or, 9x2 – 180x + 25x – 500 = 0

or, 9x ( x – 20 ) + 25 ( x – 20 ) = 0

or, ( x – 20 ) ( 9x + 25 ) = 0

Either, ( x – 20 ) = 0

or, x = 20

or, ( 9x + 25 ) = 0

or, 9x = -25

or, x = -25 / 9

Since time can not be in negative, therefore, the value of x should not be -25 / 9 minutes

Therefore, x = 20 minutes

Required time taken to fill the reservoir separately by first pipe is 20 minutes and by second pipe is ( 20 + 5 ) = 25 mintues.

8 ) Porna and Pijush together complete a work in 4 days. If they work separately, then the time taken by Porna would be 6 days more than the time taken by Pijush. Calculate the time taken by Porna alone to complete the work.

Solution : Let time taken to complete the work by Porna alone is ‘ x ‘ days

Time taken to complete the work by PIjush alone = ( x – 6 ) days

For Porna :

In x days, the part of work is completed = 1

In 1 day, ” ” ” ” ” ” ” ” ” ” ” ” = 1 / x part

For Pijush :

In ( x – 6 ) days, the part of work is completed = 1

In 1 day, ” ” ” ” ” ” ” ” ” ” ” ” = 1 / ( x – 6 ) part

For both :

In 1 day, the part of the work is completed = 1 / x + 1 / ( x – 6 )

= ( x – 6 + x ) / x ( x – 6 )

= ( 2x – 6 ) / ( x2 – 6x )

Time required to complete ( 2x – 6 ) / ( x2 – 6x ) parts of work = 1 day

Time required to complete one work = 1 / { 2x – 6 ) / ( x2 – 6x ) } days

= ( x2 – 6x ) / ( 2x – 6 ) days

According to problem

or,( x2 – 6x ) / ( 2x – 6 ) = 4

or, x2 – 6x = 8x – 24

or, x2 -6x – 8x + 24 = 0

or, x2 – 14x + 24 = 0

or, x2 – ( 12 + 2 ) x + 24 = 0

or, x2 – 12x – 2x + 24 = 0

or, x ( x – 12 ) – 2 ( x – 12 ) = 0

or, ( x – 12 ) ( x – 2 ) = 0

Either, ( x – 12 ) = 0

or, x = 12

or, ( x – 2 ) = 0

or, x = 2

Now, if time taken to complete the work by Porna alone is 2 days then time taken to complete the work by Pijush alone will be ( x – 6 ) = ( 2 – 6 ) = – 4 days which is never possible.

Hence, time taken to complete the work by Porna alone is 12 days.

9 ) If the price of 1 dozen pen is reduced by Rs. 6, then 3 more pens wil be got for Rs. 30. Before the reduction of price, calculate the price of 1 dozen pen.

Solution :

Before reduction of the price :

Let the price of 1 dozen pen = Rs. x

In Rs. x, number of pens available = 12

In Rs. 1, ” ” ” ” ” ” ” ” ” ” ” ” ” = 12 / x

In Rs. 30, ” ” ” ” ” ” ” ” ” ” = 360 / x

After reduction of the price :

The price of 1 dozen pens = Rs. ( x – 6 )

In Rs. ( x – 6 ), number of pens available = 12

In Rs. 1, ” ” ” ” ” ” ” ” ” ” ” ” = 12 / ( x – 6 )

In Rs. 30, ” ” ” ” ” ” ” ” ” ” ” ” = 360 / ( x – 6 )

According to problem

360 / ( x – 6 ) – 360 / x = 3

or, 360 { 1 / ( x – 6 ) – 1 / x} = 3

or, 120 [ { ( x – ( x – 6 ) } / x ( x – 6 ) ] = 1

or, 120 [ ( x – x + 6 ) / ( x2 – 6x ) ] = 1

or, 120 * 6 / ( x2 – 6x ) = 1

or, 720 / ( x2 – 6x ) = 1

or, x2 – 6x = 720

or, x2 – 6x – 720 = 0

or, x2 – ( 30 – 24 )x – 720 = 0

or, x2 – 30x + 24x – 720 = 0

or, x ( x – 30 ) + 24 ( x – 30 ) = 0

or, ( x – 30 ) ( x + 24 ) = 0

Either, ( x – 30 ) = 0

or, x = 30

or, ( x + 24 ) = 0

or, x = – 24

Since, the price of any thing can nor be in negative, so the value of x should not be – 24.

Therefore, the price of 1 dozen pens before reduction of the price is Rs. 30

10 ) Sathi has drawn a right angled triangle whose length of of the hypotenuse is 6 cm more than the twice of that of the shortest side. If the length of the third side is 2 cm less than the length of the hypotenuse,then calculate the lengths of the three sides of the right angled triangle drawn by Sathi. ( Solve by applying Sridhar Acharya Formula )

Solution :

Let the length of the shortest side = base ( b ) = x cm

Hypotenuse ( h ) = ( 2x + 6 ) cm

Perpendicular ( p ) = h – 2 = 2x + 6 – 2 = 2x + 4

From Pythagoras Theorem, we have

h2 = p2 + b2

or, ( 2x + 6 )2 = ( 2x + 4 )2 + x2

or, ( 2x )2 + ( 6 )2 + 2 ( 2x ) ( 6 ) = ( 2x ) 2 + 42 + 2 ( 2x ) ( 4 ) + x2

or, 4x2 + 36 +24 x = 4x2 + 16 + 16x + x2

or, -x2 + 24x – 16x + 36 – 16 = 0

or, – x2 + 8x + 20 = 0

or, – ( x2 – 8x – 20 ) = 0

or, x2 – 8x – 20 = 0

By comparing above equation with the general form of the quadratic equation ax2 + bx + c = 0 , we get

a = 1, b = – 8 , c = -20

Discriminant ( D ) = ( b2 – 4 ac ) = ( – 8 )2 – 4 ( 1 ) ( -20 ) = 64 + 80 = 144

From Sridhar Acharya’s Formula, we have

x = { – b ±  ( b2 – 4 ac ) } / 2a

or, x = { – ( -8 ) ±  144 } / 2 ( 1 )

or, x = { 8 ± 12 } / 2

Either, x = ( 8 + 12 ) / 2 = 20 / 2 = 10

or, x = ( 8 – 12 ) / 2

or, x = -4 / 2

or, x = -2

The lenght of any side of a triangle can never be in negative.

Therefore, the value of x should be 10cm

Hence, the lenght of the shortest side = 10 cm

The lenght of the hypotenuse = 2x + 6 = 2 ( 10 ) + 6 = 20 + 6 = 26 cm

The length of the third side = 2x + 4 = 2 ( 10 ) + 4 = 20 + 4 = 24 cm

11 ) If a two digit positive number is multiplied by its unit digit, then the product is 189 and if the tens digit is twice the unit digit, calculate the unit digit.( Solve by applying Sridhar Acharya Formula or by factoirsation method )

Solution : Let the unit digit = x

Tens digit = 2x

Number = { 10 * 2x } + x = 20x + x = 21 x

According to problem,

21x * x = 189

or, x2 = 189 / 21

or, x2 = 9

or, x2 – 32= 0

or, ( x + 3 ) ( x – 3 ) = 0

Either, ( x + 3 ) = 0

or, x = – 3

or, ( x – 3 ) = 0

or, x = 3

since, the number is positive, the value of x should not be -3

Hence, the required unit digit be 3.

12 ) The speed of Salma is 1 m / sec more than the speed of Anik. In a 180 m race, Salma reaches 2 seconds before than Anik. Find out the speed of Anik in m / sec .( Solve by applying Sridhar Acharya Formula or with the help of Factorisation )

Solution: For Anik :

Speed ( v ) = x m/s

Distance ( s ) = 180 m

Time ( t ) = s / v = 180 / x sec

For Salma :

Speed ( v ) = ( x + 1 ) m/s

Distance ( s ) = 180 m

Time ( t ) = s / v = 180 / ( x + 1 ) sec

According to problem

180 / x – 180 / ( x + 1 ) = 2

or, 180 { 1 / x – 1 / ( x + 1 ) } = 2

or, 90 { ( x + 1 – x ) / x ( x + 1 ) } = 1

or, 90 / ( x2 + x ) = 1

or, x2 + x = 90

or, x2 + x – 90 = 0

or, x2 + ( 10 – 9 ) x – 90 = 0

or, x2 + 10 x – 9x – 90 = 0

or, x( x + 10 ) – 9 ( x + 10 ) = 0

or, ( x – 10 ) ( x – 9 ) = 0

Either, ( x +10 ) = 0

or, x = – 10

Or, ( x – 9 ) = 0

or, x = 9

Since the value of speed can not be in negative, therefore, the value of x should not be – 10

Hence, the required speed of Anik is 9 m/s.

13 ) There is a square park in our locality. The area of a rectangular park is 78 sqm less than the twice of the area of that square-shaped park whose length si 5 m more than the length of the side of that park and the breadth is 3 m less than the length of the side of that park. Find the length of the side of the square-shaped park. ( Solve by applying Sridhar Acharya Formula or with the help of Factorisation )

Solution :

Let lenght of the square shaped park = x m

Area = x2 sqm

For rectangular park :

Length ( l ) = x + 5

breadth ( b ) = x – 3

Area = l * b = ( x + 5 ) ( x – 3 )

= x2 + 5x – 3x – 15

= x2 + 2x – 15

According to problem

2x2 – ( x2 + 2x – 15 ) = 78

or, 2x2 – x2 – 2x + 15 – 78 = 0

or, x2 – 2x – 63 = 0

or, x2 – ( 9 – 7 )x – 63 = 0

or, x2 – 9x + 7x – 63 = 0

or, x ( x – 9 ) + 7 ( x – 9 ) = 0

or, ( x – 9 ) ( x + 7 ) = 0

Either, x – 9 = 0

or, x = 9

or, x + 7 = 0

or, x = -7

Since the length of the square shaped park can never be in negative, therefore the value of x should not be – 7

Hence, the required length of the square shaped park is 9 m

14 ) In a village, Proloy babu bought 350 chilli plants for planting in his rectangular land. When he put the plants in rows, he noticed that if he would put 24 plants more than the number of rows in each row, 10 plants would remain excess. Calculate the number of rows.( Solve by applying Sridhar Acharya Formula or with the help of Factorisation )

Solution : Let the number of rows be x

Number of plants in each row = x + 24

Total number of plants that would be put in the land = x ( x + 24 ) = x2 + 24

Excess plants = 10

According to problem

x2 + 24 x+ 10 = 350

or, x2 + 24x + 10 – 350 = 0

or, x2 + 24x – 340 = 0

or, x2 + ( 34 -10 ) x – 340 = 0

or, x2 + 34x – 10x – 340 = 0

or, x ( x + 34 ) – 10 ( x + 34 ) = 0

or, ( x + 34 ) ( x – 10 ) = 0

Either, ( x + 34 )= 0

or, x + 34 = 0

or x = -34

or (x – 10 ) = 0

or, x = 10

The number of rows can never be in negative, therefore, the value of x should not be -34

The required number of rows is 10

15 ) Joseph and Kuntal work in a factory. Joseph takes 5 minutes less time than Kuntal to make a product. Joseph makes 6 products more than Kuntal while working for 6 hours. Find the number of products Kuntal makes during that time.( Solve by applying Sridhar Acharya Formula or with the help of Factorisation )

Solution : Let number of products made by Kuntal in 6 hours = x

Number of products made by Joseph in 6 hours = x + 6

To make x product by Kuntal time required = 6 hours

To make a product by Kuntal, time required = 6 / x hours

Similarly, time required to make a product by Joseph = 6 / ( x + 6 ) hours

According to problem

6 / ( x + 6 ) – 6 / x = 5 / 60

or, 6 {1 / x- 1 / ( x + 6 ) } = 5 / 60

or, 6 { ( x + 6 ) – x } / x ( x + 6 ) = 5 / 60

or, 6 { x + 6 – x} / ( x2 + 6x ) = 5 / 60

or, 36 / ( x2 + 6x ) = 1 / 12

or, x2 + 6x = 72

or, x2 + 6x – 72 = 0

or, x 2 + ( 12 – 6 ) x – 72 = 0

or, x2 + 12x – 6x – 72 = 0

or, x ( x + 12 ) – 6 ( x – 12 ) = 0

or, ( x + 12 ) ( x – 6 ) = 0

Either, ( x + 12 ) = 0

or, x = -12

or, ( x – 6 ) = 0

or, x = 6

Since, the number of product can not be negative, the value of x should not be -12

The required number of product made by Kuntal = 6

16 ) The speed of a boat in still water is 8 km / hr. If the boat can go 15 km down stream and 22km up stream in 5 hours , then calculate the speed of the stream. ( Solve by applying Sridhar Acharya Formula or with the help of Factorisation )

Soluiton : Let the speed of the stream = x km / hr

For downstream :

speed ( v ) = ( 8 + x ) km /hr

distance (s ) = 15 km

tiem = s / v = 15 / ( 8 + x) hr

For upstream :

Speed ( v ) = ( 8 – x ) km / hr

distance ( s ) = 22 km

time = s / v = 22 / ( 8 – x ) hr

According to problem

{ 15 / ( 8 + x ) }+ { 22 / ( 8 -x ) } = 5

or, { 15 ( 8 – x ) + 22 ( 8 + x ) } / ( 8 – x ) ( 8 + x ) = 5

or, ( 120 – 15x + 22x + 176 ) / ( 82 – x2 ) = 5

or, ( 7x + 296 ) / ( 64 – x2 ) = 5

or, 7x + 296 = 320 – 5x2

or, 5x2 + 7x + 296 – 320 = 0

or, 5x2 + 7x – 24 = 0

or, 5x2 + ( 15 – 8 )x – 24 = 0

or, 5x2 + 15x – 8x – 24 = 0

or, 5x ( x + 3 ) – 8 ( x + 3 ) = 0

or, ( x + 3 ) ( 5x – 8 ) = 0

Either,

( x + 3 ) = 0

or, x = -3

or, ( 5x – 8 ) = 0

or, 5x = 8

or, x = 8 / 5 = 1.6

Since, the speed can not be negative, the value of x should not be -3

Hence, the speed of the stream is 1.6 km / hr

17 ) A superfast train runs at the speed of 15 km / hr more than that of an express train. Leaving the same station the superfast train reached a station of 180 kms. distance 1 hour before than the express train. Determine the speed of the superfast train in km / hr.( Solve by applying Sridhar Acharya Formula or with the help of Factorisation )

Solution :

For superfast train :

let speed ( v ) = x km / hr

distance ( s ) = 180 km

time = s / v = 180 / x hr

For express train:

speed ( v ) = ( x – 15 ) km / hr

distance ( s ) = 180 km

time = s / v = 180 / ( x – 15 ) hr

According to problem

180 / ( x – 15 ) – 180 / x = 1

or , 180 { 1 / ( x – 15 ) – 1 / x } = 1

or, 180 { x- ( x – 15 ) } / x ( x – 15 ) = 1

or, 180 ( x – x + 15 ) / ( x2 – 15x ) = 1

or, 2700 / ( x2 – 15x ) = 1

or, x2 – 15x = 2700

or, x2 – 15x – 2700 = 0

or,, x2 – ( 60 – 45 ) x – 2700 = 0

or, x2 – 60x + 45x – 2700 = 0

or, x( x – 60 ) + 45 ( x – 60 ) = 0

or, ( x – 60 ) ( x + 45 ) = 0

Either, ( x – 60 ) = 0

or, x = 60

or ( x + 45 ) = 0

or, x = – 45

Since, the value of speed can not be negative, the value of x should not be -45

Therefore, the required speed of the superfast train = 60 km / hr

18 ) Rehana went to the market and saw that the price of dal of 1 kg is Rs. 20 and the price of rice of 1 kg is Rs. 40 less than that of price of 1 kg fish. The total quantity of fish and that of dal each in Rs. 240 is equal to the quantity of rice in Rs 280. Caculate the cost price of 1 kg. fish.( Solve by applying Sridhar Acharya Formula or with the help of Factorisation )

Solution :

Let the price of the 1 kg fish = Rs. x

Price of 1 kg of rice = Rs ( x – 40 )

Price of 1 kg dal = Rs. ( x – 20 )

Quantity of Fish available in Rs. 240 = 240 / x

Quantity of Dal available in Rs. 240 = 240 / ( x – 20 )

Quantity of Rice available in Rs. 280 = 280 / ( x – 40 )

According to problem

{ 240 / x } + { 240 / ( x – 20 ) } = 280 / ( x – 40 )

or, 240 { ( x – 20 + x ) / x ( x – 20 ) } = 280 / ( x – 40 )

or, 6 ( 2x – 20 ) / ( x2 – 20x ) = 7 / ( x – 40 )

or, ( 12x – 120 ) / ( x2 – 20x ) = 7 / ( x – 40 )

or, ( 12x- 120 ) ( x – 40 ) = 7 ( x2 – 20x )

or, 12x2 – 120x – 480x + 4800 = 7x2 – 140x

or, 12x2 – 7x2 – 600x + 140x + 4800 = 0

or, 5x2 – 460x + 4800 = 0

or, 5 ( x2 – 92x + 960 ) = 0

or, x2 – 92x + 960 = 0

or, x2 – ( 80 + 12 )x + 960 = 0

or, x2 – 80x – 12x + 960 = 0

or, x ( x – 80 ) – 12 ( x – 80 ) = 0

or, ( x – 80 ) ( x – 12 ) = 0

Either ( x- 80 ) = 0

or, x = 80

Or, ( x – 12 ) = 0

or, x = 12

If the cost price of 1 kg fish be Rs. 12 then the price of dal and rice would be ( 12 – 20 ) = Rs. – 8 and ( 12 – 40 ) = Rs. – 28 respectively bu the price can not be negative. Therfore, the value of x should not be 12.

The required price of 1 kg fish = Rs. 80

Type – 03

1 ) Solve : 1 / x + 1 / ( x – 5 ) = 1 / 6

Solution : 1 / x + 1 / ( x – 5 ) = 1 / 6

The required solutions are 15 or 2.

2 ) Solve : 9x2 – 155x – 500 = 0

Solution : 9x2 – 155x – 500 = 0

or, 9x2 – ( 180 – 25 ) x – 500 = 0

or, 9x2 – 180x + 25x – 500 = 0

or, 9x ( x – 20 ) + 25 ( x – 20 ) = 0

or, ( x – 20 ) ( 9x + 25 ) = 0

Either ( x – 20 ) = 0

or, x = 20

or, ( 9x + 25 ) = 0

or, 9x = -25

or, x = – 25 / 9

The required solution of x are 20 and -25 / 9

3 ) 3y2 – 20 = 160 – 2y2

or, 3y2 + 2y2 = 160 + 20

or, 5y2 = 180

or, y2 = 36

or, y2 – 62 = 0

or, ( y + 6 ) ( y – 6 ) = 0

Either, ( y + 6 ) = 0

or, y = -6

Or, ( y – 6 ) = 0

or, y = 6

The required solutions are 6 and – 6.

4 ) Solve : ( 2x + 1 )2 + ( x + 1 )2 = 6x + 47

Solution : ( 2x + 1 )2 + ( x + 1 )2 = 6x + 47

or, ( 2x )2 + 12 + 2 ( 2x ) ( 1 ) + x2 + 2x + 1 = 6x + 47

or, 4x2 + 1 + 4x + x2 + 2x + 1 = 6x + 47

or, 5x2 + 6x – 6x = 47 – 2

or, 5x2 = 45

or, x2 = 9

or, x2 – 32 = 0

or, ( x + 3 ) ( x – 3 ) = 0

either ( x + 3 ) = 0

or, x = -3

or, ( x – 3 ) = 0

or, x = 3

The required solutions are 3 and – 3

5 ) Solve : ( x – 7 ) ( x – 9 ) = 195

Solution : ( x – 7 ) ( x – 9 ) = 195

or, x ( x – 9 ) – 7 ( x – 9 ) = 195

or, x2 – 9x – 7x + 63 – 195 = 0

or, x2 – 16x – 132 = 0

or, x2 – ( 22 – 6 )x – 132 = 0

or, x2 – 22x + 6x – 132 = 0

or, x ( x – 22 ) + 6 ( x – 22 ) = 0

or, ( x- 22 ) ( x + 6 ) = 0

Either, ( x – 22 ) = 0

or, x = 22

Or, ( x + 6 ) = 0

or, x = -6

The required solutions are 22 and -6.

6 ) 3x – 24 / x = x / 3 ( x ≠ 0 )

Solution :

7 ) x / 3 + 3 / x = 15 / x ; ( x ≠  0 )

Solution :

8 ) 10x – 1 / x = 3 ; ( x ≠  0 )

Solution :

9 ) 2 / x2 – 5/ x + 2 = 0 ; ( x ≠  0 )

Solution :

10 ) { ( x – 2 ) / ( x + 2 ) } + 6 { ( x – 2 ) / ( x – 6 ) } = 1 ; ( x ≠ -2, 6 )

Solution :

11 ) { 1 / ( x – 3 ) } – { 1 / ( x + 5 ) } = 1 / 6 ; ( x ≠ 3, -5 )

Solution :

12 ) x / ( x + 1 ) + ( x + 1 ) / x = 2 1/12 ; ( x ≠ 0 , -1 )

Solution :

13 ) { ( ax + b ) / ( a + bx ) } = { ( cx + d ) / ( c + dx ) } ; ( a ≠ b , c ≠ d and x ≠ – a/b, -c/d )

14 ) ( 2x + 1 ) + 3 / ( 2x + 1 ) = 4 , ( x ≠ – 1 / 2 )

15 ) ( x + 1 ) / 2 + 2 / ( x + 1 ) = ( x + 1 ) / 3 + 3 / ( x + 1 ) – 5 / 6 , ( x ≠ -1 )

16 ) { ( 12x + 17 ) / ( 3x + 1 ) } – { ( 2x + 5 ) / ( x + 7 ) } = 3 1 / 5 ; ( x ≠ – 1 / 3 , -7 )

17 ) {( x + 3 ) / ( x – 3 )} + 6 {( x – 3 ) / ( x + 3 )} = 5 ; ( x ≠ 3 , -3 )

18 ) 1 / ( a + b + x ) = 1 / a + 1 / b + 1 / x ; ( x ≠ 0, – ( a + b ) )

19 ) {( x + a ) / ( x – a )}2 -5{ ( x + a ) / ( x – a )} + 6 = 0 ; ( x ≠ a )

20 ) 1 / x – 1 / ( x + b ) = 1 / a – 1 / ( a + b ) , ( x ≠ 0 , -b )

21 ) { 1 / ( x – 1 )( x – 2 ) } + { 1 / ( x – 2 )( x – 3 )} + { 1 / ( x – 3 )( x – 4 ) } = 1 / 6 ; ( x ≠ 1 , 2 , 3 , 4 )

22 ) a / ( x – a ) + b / ( x – b ) = 2c / ( x – c ) , ( x ≠ a , b, c )

23 ) x2 – ( 3 + 2 ) x + 23 = 0

24 ) x2 + 7x + 12 = 0

25 ) x2 – x – 12 = 0

26 ) x2 + 9x + 8 = 0

27 ) x2 – 2x – 8 = 0

28 ) x2 + 7x – 8 = 0

29 ) 84/ x – 84 / ( x + 5 ) = 5

30 ) x2 + 5x – 204 = 0

31 ) 3x2 + x – 10 = 0

32 ) 6x2 – x – 2 = 0

33 ) 25x2 – 20x + 4 = 0

34 ) x2 + 5x + 4 = 0

34 ) 5x2 + 23x + 12 = 0

35 ) x2 + 2ax + ( a + b) ( a – b )

36) ax2 – ( a2 + 1 ) x + a = 0

37 ) 99x2 – 202 x + 99 = 0

38 ) ax2 – ( a2 – 2 ) x – 2a = 0

39 ) x2 + 1 – 6 / x2 = 0

40 ) 4 / ( x – b ) + b / ( x – a ) = 2

41 ) a / ( ax -1 ) + b / ( bx – 1 ) = a + b ; ( x ≠ 1 / a , 1 / b )

42 ) ( x + 4 ) ( 2x – 3 ) = 6

43 ) 2x2 – 6x + 1 = 0 , Solve by the method o fcompleting the square.

44 ) 9x2 + 7x – 2 = 0 , Solve by the method of factorisation.

45 ) x2 – 6x + 4 = 0 , solve wiht the help of Sridhar Acharya Formula.

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