Chapter 8: Quadratic Equations Exercise – 8.6

Question: 1Determine the nature of the roots of the following quadratic equations.

(i) 2x

^{2}– 3x + 5 = 0(ii) 2x

^{2}- 6x + 3 = 0(iii) For what value of k (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0 is a perfect square.(iv) Find the least positive value of k for which the equation x

^{2 }+ kx + 4 = 0 has real roots.(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.

Kx

^{2 }+ 2x + 1 = 0(vi) Kx

^{2}+ 6x + 1 = 0(vii) x

^{2 }- kx + 9 = 0

Solution:The given quadratic equation is in the form of ax

^{2 }+ bx + c = 0So a = 2, b = – 3, c = 5

We know, determinant (D) = b

^{2}- 4ac = (-3)^{2}- 4(2)(5) = 9 - 40 = – 31 < 0Since D < 0, the determinant of the equation is negative, so the expression does not having any real roots.

(ii) 2x

^{2}- 6x + 3 = 0

The given quadratic equation is in the form of ax^{2 }+ bx + c = 0So a = 2, b = -6, c = 3

We know, determinant (D) = b

^{2}- 4ac = (- 6)^{2}- 4(2)(3) = 36 - 24 = 12 < 0Since D > 0, the determinant of the equation is positive, so the expression does having any real and distinct roots

(iii) For what value of k (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0 is a perfect square.The given equation is (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0Here, a = 4 - k, b = 2k + 4, c = 8k + 1

The discriminate (D) = b

^{2 }- 4ac = (2k+4)^{2}- 4(4 - k)(8k + 1)= (4k

^{2 }+ 16 + 16k) - 4(32k + 4 - 8k^{2 }- k)= 4(k

^{2}+ 8k^{2 }+ 4k - 31k + 4-4)= 4(9k

^{2 }- 27k)D = 4(9k

^{2 }- 27k)The given equation is a perfect square D = 0

4(9k

^{2 }- 27k) = 09k

^{2 }- 27k = 0Taking out common of 3 from both sides and cross multiplying = k

^{2 }- 3k = 0= K (k - 3) = 0

Either k = 0 Or k = 3

The value of k is to be 0 or 3 in order to be a perfect square.

(iv)Find the least positive value of k for which the equation x

^{2 }+ kx + 4 = 0 has real roots.The given equation is x

^{2 }+ kx + 4 = 0 has real roots Here, a = 1, b = k, c = 4The discriminate (D) = b

^{2 }- 4ac = 0 = k^{2}- 16 = 0 = k = 4, k = – 4The least positive value of k = 4 for the given equation to have real roots.

(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.

Kx^{2 }+ 2x + 1 = 0The given equation is Kx

^{2 }+ 2x + 1 = 0Here, a = k, b = 2, c = 1

The discriminate (D) = b

^{2 }- 4ac = 0= 4 - 4k = 0

= 4k = 4

K = 1

The value of k = 1 for which the quadratic equation is having real and equal roots.

(vi) Kx^{2}+ 6x + 1 = 0The given equation is Kx

^{2 }+ 6x + 1 = 0Here, a = k, b = 6, c = 1

The discriminate (D) = b

^{2 }- 4ac = 0= 36 - 4k = 0

= 4k = 36

K = 9

The value of k = 9 for which the quadratic equation is having real and equal roots.

(vii) x

^{2 }- kx + 9 = 0The given equation is x

^{2 }- kx + 9 = 0Here, a = 1, b = – k, c = 9

Given that the equation is having real and distinct roots.

Hence, the discriminate (D) = b

^{2 }- 4ac = 0= k

^{2}- 4(1)(9) = 0= k

^{2 }– 36 = 0= K = – 6 and k = 6

The value of k lies between -6 and 6 respectively to have the real and distinct roots.

Question: 2Find the value of k (i) Kx

^{2 }+ 4x + 1 = 0.(ii)

(iii) 3x

^{2 }- 5x + 2k = 0(iv) 4x

^{2 }+ kx + 9 = 0(v) 2kx

^{2 }- 40x + 25 = 0(vi) 9x

^{2 }- 24x + k = 0(vii) 4x

^{2}- 3kx + 1 = 0(viii) x

^{2 }- 2(5 + 2k)x + 3(7 + 10k) = 0(ix) (3k +1)x

^{2}+ 2(k +1)x + k = 0(x) Kx

^{2 }+ kx + 1 = – 4x^{2 }- x(xi) (k + 1)x

^{2 }+ 2(k + 3)x + k + 8 = 0(xii) x

^{2 }- 2kx + 7k - 12 = 0(xiii) (k + 1)x

^{2 }- 2(3k + 1)x + 8k + 1 = 0(xiv) 5x

^{2 }- 4x + 2 + k(4x^{2 }- 2x + 1) = 0(xv) (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0(xvi) (2k + 1)x

^{2 }+ 2(k + 3)x + (k +5 ) = 0(xvii) 4x

^{2 }- 2(k + 1)x + (k + 4) = 0

Solution:The given equation Kx

^{2 }+ 4x + 1 = 0 is in the form of ax^{2 }+ bx + c = 0Where a = k, b = 4, c = 1

Given that, the equation has real and equal roots D = b

^{2 }- 4ac = 0= 4

^{2 }- 4(k)(1) = 0= 16 – 4k = 0

= k = 4

The value of k is 4

(ii)

The given equationis in the form of ax

^{2 }+ bx + c = 0 where a= k, b = - 2√5, c = 4.Given that, the equation has real and equal roots D = b

^{2}- 4ac = 020 - 16k = 0

K = 5 /4

The value of k is k = 5/4

(iii) 3x

^{2 }- 5x + 2k = 0The given equation 3x

^{2 }- 5x + 2k = 0 is in the form of ax^{2 }+ bx + c = 0 where a = 3, b = – 5, c = 2kGiven that, the equation has real and equal roots D = b

^{2 }- 4ac = 0= (- 5)

^{2 }- 4(3)(2k) = 0= 25 - 24k = 0

K = 25/24

The value of the k is k = 25/24

(iv) 4x

^{2 }+ kx + 9 = 0The given equation 4x

^{2 }+ kx + 9 = 0 is in the form of ax^{2 }+ bx + c = 0 where a = 4, b = k, c = 9Given that, the equation has real and equal roots D = b

^{2 }- 4ac = 0= k

^{2 }- 4(4)(9) = 0= k

^{2 }- 144 = 0= k = 12

The value of k is 12

(v) 2kx

^{2 }- 40x + 25 = 0The given equation 2kx

^{2 }- 40x + 25 = 0 is in the form of ax^{2 }+ bx + c = 0 where a = 2k, b = – 40, c = 25Given that, the equation has real and equal roots D = b

^{2 }- 4ac = 0(-40)

^{2 }- 4(2k)(25) = 01600 - 200k = 0

k = 8

The value of k is 8

(vi) 9x

^{2 }- 24x + k = 0The given equation 9x

^{2 }- 24x + k = 0 is in the form of ax^{2 }+ bx + c = 0 where a = 9, b = – 24, c = kGiven that, the equation has real and equal roots D = b

^{2 }- 4ac = 0( – 24)

^{2 }- 4(9)(k) = 0576 - 36k = 0

k = 16

The value of k is 16

(vii) 4x

^{2}- 3kx + 1 = 0The given equation 4x

^{2 }- 3kx + 1 = 0 is in the form of ax^{2 }+ bx + c = 0 where a = 4, b = - 3k, c = 1Given that, the equation has real and equal roots D = b

^{2 }- 4ac = 0= (-3k)

^{2 }- 4(4)(1) = 0= 9k

^{2 }- 16 = 0K = 4/3

The value of k is 4/3

(viii) x

^{2 }- 2(5 + 2k)x + 3(7 + 10k) = 0The given equation X

^{2 }- 2(5 + 2k)x + 3(7 + 10k) = 0 is in the form of ax^{2 }+ bx + c = 0 where a = 1, b = +2(52k), c = 3(7 + 10k)Given that, the nature of the roots of the equation are real and equal roots D = b

^{2 }- 4ac = 0= (+2(52k))

^{2 }- 4(1)(3(7 + 10k)) = 0= 4(5 + 2k)

^{2 }- 12(7 + 10k) = 0= 25 + 4k

^{2 }+ 20k - 21 - 30k = 0= 4k

^{2 }- 10k + 4 = 0Simplifying the above equation.

We get, = 2k

^{2 }- 5k + 2 = 0= 2k

^{2 }- 4k - k + 2 = 0= 2k(k - 2) - 1(k - 2) = 0

= (k - 2)(2k - 1) = 0, K = 2 and k = 1/2 The value of k can either be 2 or 1/2

(ix) (3k +1)x^{2}+ 2(k +1)x + k = 0The given equation (3k + 1)x

^{2 }+ 2(k + 1)x + k = 0 is in the form of ax^{2 }+ bx + c = 0 where a = 3k + 1, b = +2(k + 1), c = (k)Given that, the nature of the roots of the equation are real and equal roots D = b

^{2 }- 4ac = 0= [2(k + 1)]

^{2 }- 4(3k + 1)(k) = 0= (k + 1)

^{2 }- k(3k + 1) = 0= -2k

^{2 }+ k + 1 = 0This equation can also be written as 2k

^{2 }- k - 1 = 0The value of k can be obtained by k

The value of k are 1 and (-1)/2 respectively.

(x) Kx^{2 }+ kx + 1 = -4x^{2 }- xBringing all the x components on one side we get, x

^{2}(4 + k) + x(k + 1) + 1 = 0The given equation Kx

^{2 }+ kx + 1 = -4x^{2 }- x is in the form of ax^{2 }+ bx + c = 0 where a = 4 + k,b = +k + 1, c = 1 Given that, the nature of the roots of the equation are real and equal roots D= b^{2}- 4ac = 0= (k+1)

^{2}- 4(4 + k)(1) = 0= k

^{2}- 2k – 10 = 0The equation is also in the form ax

^{2 }+ bx + c = 0The value of k is obtained by a = 1, b = -2, c = – 15

Putting the respective values in the above formula we will obtain the value of k

The value of k are 5 and -3 for different given quadratic equation.

(xi) (k + 1)x

^{2 }+ 2(k + 3)x + k + 8 = 0The given equation (k + 1) x

^{2 }+ 2(k + 3)x + k + 8 = 0 is in the form of ax^{2 }+ bx + c = 0 where a = k + 1,b = 2(k + 3), c = k + 8Given the nature of the roots of the equation are real and equal. D = b

^{2 }- 4ac = 0= [2(k + 30]

^{2 }- 4(k + 1)(k + 8) = 0= 4(k + 3)

^{2 }- 4(k + 1)(k + 8) = 0Taking out 4 as common from the LHS of the equation and dividing the same on the RHS = (k + 3)

^{2}-(k + 1)(k + 8) = 0= k

^{2 }+ 9 + 6k - (k^{2 }+ 9k + 18) = 0Cancelling out the like terms on the LHS side = 9 + 6k - 9k - 8 = 0

= - 3k + 1 = 0

= 3k = 1

K = 1/3

The value of k of the given equation is k =1/3

(xii) x

^{2 }- 2kx + 7k - 12 = 0The given equation is x

^{2 }- 2kx + 7k - 12 = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 where a = 1, b = – 2k, c = 7k – 12Given the nature of the roots of the equation are real and equal. D = b

^{2 }- 4ac = 0= (2k)

^{2 }- 4(1)(7k - 12) = 0= 4k

^{2 }- 28k + 48 = 0= k

^{2 }- 7k + 12 = 0The value of k can be obtained by

Here a = 1, b = – 7k, c = 12

By calculating the value of k is

The value of k for the given equation is 4 and 3 respectively.

(xiii) (k + 1)x

^{2 }- 2(3k + 1)x + 8k + 1 = 0The given equation is (k + 1)x

^{2 }- 2(3k + 1)x + 8k + 1 = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 where a = k + 1, b = – 2(k + 1), c = 8k + 1Given the nature of the roots of the equation are real and equal. D = b

^{2}- 4ac = 0= (-2(k + 1))

^{2 }- 4(k + 1)(8k + 1) = 0= 4(3k + 1)

^{2 }- 4(k + 1)(8k + 1) = 0Taking out 4 as common from the LHS of the equation and dividing the same on the

RHS = (3k + 1)

^{2 }- (k + 1)(8k + 1) = 0= 9k

^{2 }+ 6k + 1 – (8k^{2 }+ 9k + 1) = 0= 9k

^{2 }+ 6k + 1 – 8k^{2 }- 9k - 1 = 0= k

^{2 }- 3k = 0= k(k - 3) = 0

Either k = 0 Or, k - 3 = 0 = k = 3

The value of k for the given equation is 0 and 3 respectively.

(xiv) 5x

^{2 }- 4x + 2 + k(4x^{2 }- 2x + 1) = 0The given equation 5x

^{2 }- 4x + 2 + k(4x^{2 }- 2x + 1) = 0 can be written as x^{2}(5 + 4k) - x(4 + 2k) + 2 - k = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 where a = 5 + 4k, b = -(4 + 2k), c = 2 - kGiven the nature of the roots of the equation are real and equal.

D = b

^{2 }- 4ac = 0= [-(4 + 2k)]

^{2 }- 4(5 + 4k)(2 - k) = 0= 16 + 4k

^{2 }+ 16 - 4(10 - 5k + 8k - 4k^{2}] = 0= 16 + 4k

^{2 }+ 16 - 40 + 20k - 32k + 16k^{2}= 0= 20k

^{2 }- 4k - 24 = 0Taking out 4 as common from the LHS of the equation and dividing the same on the RHS = 5k

^{2 }- k - 6 = 0 The value of k can be obtained by equationThe value of k for the given equation are k = 6/5 and−1 respectively.

(xv) (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0The given equation is (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 where a = 4 - k, b = (2k + 4), c = 8k + 1Given the nature of the roots of the equation are real and equal.

D = b

^{2 }- 4ac = 0= (2k + 4)

^{2 }- 4(4 - k)(8k + 1) = 0= 4k

^{2 }+ 16k + 16 - 4(-8k^{2 }+ 32k + 4 - k) = 0= 4k

^{2 }+ 16k + 16 + 32k^{2 }- 124k - 16 = 0Cancelling out the like and opposite terms.

We get, = 36k

^{2 }- 108k = 0Taking out 4 as common from the LHS of the equation and dividing the same on the RHS = 9k

^{2 }- 27k = 0= 9k(k -3 ) = 0

Either 9k = 0 K = 0 Or, k - 3 = 0 K = 3

The value of k for the given equation is 0 and 3 respectively.

(xvi) (2k + 1)x

^{2 }+ 2(k + 3)x + (k +5 )= 0The given equation is (2k + 1)x

^{2 }+ 2(k + 3)x + (k + 5) = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 where a = 2k + 1, b = 2(k + 3), c = k + 5Given the nature of the roots of the equation are real and equal.

D = b

^{2 }- 4ac = 0= [2(k + 3)]

^{2 }- 4(2k + 1)(k + 5) = 0Taking out 4 as common from the LHS of the equation and dividing the same on the

RHS = [(k + 3)]

^{2 }- (2k + 1)(k + 5) = 0= K

^{2 }+ 9 + 6k - (2k^{2 }+ 11k + 5) = 0= – k

^{2 }- 5k + 4 = 0= k

^{2 }+ 5k - 4 = 0The value of k can be obtained by k = 6/5 and − 1 respectively.

Here a = 1, b = 5, c = – 4

The value of k for the given equation is

(xvii) 4x

^{2 }- 2(k + 1)x + (k + 4) = 0The given equation is 4x

^{2 }- 2(k + 1)x + (k + 4) = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 where a = 4, b = -2(k + 1), c = k + 4Given the nature of the roots of the equation are real and equal.

D = b

^{2 }- 4ac = 0= [-2(k + 1)]

^{2 }- 4(4)(k + 4) = 0Taking out 4 as common from the LHS of the equation and dividing the same on the

RHS = (k + 1)

^{2 }- 4(k + 4) = 0= k

^{2 }+ 1 + 2k - 4k - 16 = 0= k

^{2 }- 2k - 15 = 0The value of k can be obtained by k = 6/5 and − 1 respectively.

Here a = 1, b = -2, c = -15

The value of k for the given equation is

Question: 3In the following, determine the set of values of k for which the given quadratic equation has real roots:

(i) 2x

^{2 }+ 3x + k = 0(ii) 2x

^{2 }+ kx + 3 = 0(iii) 2x

^{2 }- 5x - k = 0(iv) Kx

^{2 }+ 6x + 1= 0(v) x

^{2 }- kx + 9 = 0

Solution:(i) 2x

^{2 }+ 3x + k = 0The given equation is 2x

^{2 }+ 3x +k = 0The given quadratic equation has equal and real roots D = b

^{2}- 4ac = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = 2, b = 3, c = k = 9 - 4(2)(k) = 0= 9 - 8k = 0

= k ≤ 98

The value of k does not exceed k ≤ 98 to have a real root.

(ii) 2x

^{2 }+ kx + 3 = 0The given equation is 2x

^{2 }+ kx + 3 = 0The given quadratic equation has equal and real roots D = b

^{2 }- 4ac = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = 2, b = k, c = 3 = k^{2 }- 4(2)(3) = 0= k

^{2 }– 24 = 0The value of k should not exceedin order to obtain real roots.

(iii) 2x^{2 }- 5x - k = 0The given equation is 2x

^{2 }- 5x - k = 0The given quadratic equation has equal and real roots D = b

^{2 }- 4ac = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = 2, b = - 5, c = - k = 25 - 4(2)(- k) = 0= 25 - 8k = 0

= k ≤ 25/8

The value of k should not exceed k ≤ 25/8

(iv) Kx^{2 }+ 6x + 1= 0The given equation is Kx

^{2 }+ 6x + 1 = 0The given quadratic equation has equal and real roots D = b

^{2 }- 4ac = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = k, b = 6, c = 1 = 36 - 4(k)(1) = 0= 36 - 4k = 0

= k = 9

The value of k for the given equation is k = 9

(v) x^{2 }- kx + 9 = 0The given equation is X

^{2 }- kx + 9 = 0The given quadratic equation has equal and real roots D = b

^{2}- 4ac = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = 1, b = -k, c = 9 = k^{2 }- 4(1)(-9) = 0= k

^{2 }– 36 = 0= k

^{2 }= 36 k ≥ √36 K = 6 and k = - 6The value of k should in between K = 6 and k = - 6 in order to maintain real roots.

Question: 4

Determine the nature of the roots of the following quadratic equations.(i) 2x

^{2}- 3x + 5 = 0(ii) 2x

^{2}- 6x + 3 = 0(iii) For what value of k (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0 is a perfect square(iv) Find the least positive value of k for which the equation x

^{2 }+ kx + 4 = 0 has real roots.(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.

Kx

^{2 }+ 2x + 1 = 0(vi) Kx

^{2}+ 6x + 1 = 0(vii) x

^{2 }– kx + 9 = 0

Solution:(i) 2x

^{2}- 3x + 5 = 0The given quadratic equation is in the form of ax

^{2 }+ bx + c = 0So a = 2, b = -3, c = 5

We know, determinant (D) = b

^{2}– 4ac = (-3)^{2}– 4(2)(5)= 9 – 40 = – 31 < 0

Since D < 0, the determinant of the equation is negative, so the expression does not having any real roots.

(ii) 2x

^{2}- 6x + 3 = 0The given quadratic equation is in the form of ax

^{2 }+ bx + c = 0So a = 2, b = -6, c = 3

We know, determinant (D) = b

^{2}– 4ac = (-6)^{2}- 4(2)(3) = 36 – 24 = 12 < 0Since D > 0, the determinant of the equation is positive, so the expression does having any real and distinct roots.

(iii) For what value of k (4 - k)x^{2 }+ (2k + 4)x + (8k + 1) = 0 is a perfect squareThe given equation is (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0Here, a = 4 - k, b = 2k + 4, c = 8k + 1

The discriminate (D) = b

^{2 – }4ac= (2k + 4)

^{2}– 4(4 - k)(8k + 1)= (4k

^{2 }+ 16 + 16k) - 4(32k + 4 - 8k^{2 }- k)= 4(k

^{2 }+ 8k^{2 }+ 4k - 31k + 4 - 4)= 4(9k

^{2 }- 27k)= 4(9k

^{2 }- 27k)The given equation is a perfect square D = 0

4(9k

^{2 }- 27k) = 09k

^{2 }- 27k = 0Taking out common of 3 from both sides and cross multiplying K

^{2 }- 3k = 0 K (k - 3) = 0Either k = 0 Or k = 3

The value of k is to be 0 or 3 in order to be a perfect square.

(iv) Find the least positive value of k for which the equation x

^{2 }+ kx + 4 = 0 has real roots.The given equation is x

^{2 }+ kx + 4 = 0 has real roots Here, a = 1, b = k, c = 4The discriminate (D) = b

^{2 – }4ac = 0= k

^{2}– 16 = 0= k = 4, k = - 4

The least positive value of k = 4 for the given equation to have real roots.

(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.

Kx

^{2 }+ 2x + 1 = 0The given equation is Kx

^{2 }+ 2x + 1 = 0Here, a = k, b = 2, c = 1

The discriminate (D) = b

^{2 – }4ac = 0= 4 - 4k = 0

= 4k = 4

K = 1

The value of k = 1 for which the quadratic equation is having real and equal roots.

(vi) Kx

^{2}+ 6x + 1 = 0The given equation is Kx

^{2 }+ 6x + 1 = 0Here, a = k, b = 6, c = 1

The discriminate (D) = b

^{2 – }4ac = 0= 36 - 4k = 0

= 4k = 36

= K = 9

The value of k = 9 for which the quadratic equation is having real and equal roots.

(vii) x

^{2 }– kx + 9 = 0The given equation is X

^{2 }– kx + 9 = 0Here, a = 1, b = -k, c = 9

Given that the equation is having real and distinct roots. Hence, the discriminate (D) = b

^{2 – }4ac = 0= k

^{2}– 4(1)(9) = 0= k

^{2 }- 36 = 0= k = – 6 and k = 6

The value of k lies between -6 and 6 respectively to have the real and distinct roots.

Question: 5Find the values of k for which the given quadratic equation has real and distinct roots.

(i) Kx

^{2 }+ 2x + 1 = 0(ii) Kx

^{2 }+ 6x + 1 = 0

Solution:(i) Kx

^{2 }+ 2x + 1 = 0The given equation is Kx

^{2 }+ 2x + 1 = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = k, b = 2, c = 1 D = b^{2 }- 4ac = 0= 4 - 4(1)(k) = 0

= 4k = 4

k = 1

The value of k for the given equation is k = 1

(ii) Kx^{2 }+ 6x + 1 = 0The given equation is Kx

^{2 }+ 6x + 1 = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = k, b = 6, c = 1D = b

^{2 }- 4ac = 0= 36 - 4(1)(k) = 0

= 4k = 36

= k = 9

The value of k for the given equation is k = 9

Question: 6

For what value of k, (4 - k)x^{2 }+ (2k + 4)x + (8k + 1) = 0, is a perfect square.

Solution:The given equation is (4 - k)x

^{2 }+ (2k + 4)x + (8k + 1) = 0The given equation is in the form of ax

^{2 }+ bx + c = 0 so, a = 4 - k, b = 2k + 4, c = 8k + 1 D = b^{2 }- 4ac= (2k + 4)

^{2 }- 4(4 - k)(8k + 1)= 4k

^{2 }+ 16 + 4k - 4(32 + 4 - 8k^{2 }- k)= 4(k

^{2 }+ 4 + k - 32 - 4 + 8k^{2 }+ k)=4(9k

^{2 }- 27k)Since the given equation is a perfect square Therefore D = 0 = 4(9k

^{2 }- 27k) = 0= (9k

^{2 }- 27k) = 0= 3k(k - 3) = 0

Therefore 3k = 0

K = 0 Or, k-3 = 0 K = 3 The value of k should be 0 or 3 to be perfect square.

Question: 7

If the roots of the equation (b - c)x^{2}+ (c - a)x + (a - b) = 0 are equal , then prove that 2b = a + c.

Solution:The given equation is (b - c)x

^{2}+ (c - a)x + (a - b) = 0.The given equation is the form of ax

^{2 }+ bx + c = 0.So, a = (b - c), b = (c - a), c = (a - b)

According to question the equation is having real and equal roots.

Hence discriminant (D) = b

^{2 }- 2ac = 0= (c - a)

^{2 }- 4(b - c)(a - b) = 0= c

^{2 }+ a^{2 }- 2ac – 4(ab - b^{2 }- ac + cb) = 0= c

^{2 }+ a^{2 }- 2ac – 4ab + 4b^{2 }+ 4ac - 4cb = 0= c

^{2 }+ a^{2 }+ 2ac – 4ab + 4b^{2 }- 4cb = 0= (a + c)

^{2 }- 4ab + 4b^{2 }- 4cb = 0= (c + a - 2b)

^{2}= 0= (c + a - 2b) = 0

= c + a = 2b

Hence it is proved that c + a = 2b.

Question: 8

If the roots of the equation (a^{2 }+ b^{2}) x^{2 }– 2(ac + bd)x + (c^{2 }+ d^{2}) = 0 are equal. Prove that a ÷ b = c ÷ d.

Solution:The given equation is (a

^{2 }+ b^{2})x^{2}– 2(ac + bd)x + (c^{2 }+ d^{2}) = 0.The equation is in the form of ax

^{2 }+ bx = c = 0Hence, a = (a

^{2 }+ b^{2}), b = – 2(ac + bd), c = (c^{2 }+ d^{2}).The given equation is having real and equal roots.

Discriminant (D) = b

^{2 }- 4ac = 0= [-2(ac + bd)]

^{2}- 4 (a^{2 }+b^{2})(c^{2 }+ d^{2}) = 0= (ac + bd)

^{2}- (a^{2 }+ b^{2})(c^{2 }+ d^{2}) = 0= a

^{2}c^{2}+ b^{2}d^{2 }+ 2abcd – (a^{2}c^{2}+ a^{2}d^{2}+ b^{2}c^{2 }+ b^{2}d^{2}) = 0Cancelling out the equal and opposite terms.

We get, = 2abcd - a

^{2}d^{2}- b^{2}c^{2 }= 0= abcd + abcd - a

^{2}d^{2}- b^{2}c^{2 }= 0= ad(bc - ad) + bc(ad - bc) = 0

= ad(bc - ad) - bc(bc - ad) = 0

= (ad - bc)(bc - ad) = 0

= ad – bc = 0

= (a ÷ b) = (c ÷ d)

Hence, it is proved.

Question: 9If the roots of the equation ax

^{2 }+ 2bx + c = 0 andare simultaneously real, then prove that b^{2 }- ac = 0.

Solution:The given equations are ax

^{2 }+ 2bx +c = 0 andThese two equations are of the form ax

^{2 }+ bx + c = 0.Given that the roots of the two equations are real.

Hence, D = 0 that is b

^{2 }- 4ac = 0Let us assume that ax

^{2 }+ 2bx + c = 0 be equation (i) andbe (ii) From equation (i) b^{2 }- 4ac = 0= 4 b

^{2 }- 4ac = 0 .... (iii)From equation (ii) b

^{2 }- 4ac = 0Given, that the roots of equation (i) and (ii) are simultaneously real and hence equation (iii) = equation (iv).

= 4b

^{2 }- 4ac = 4ac - 4b

^{2}= 8ac = 8b^{2}= b^{2 }- ac = 0.Hence it is proved that b

^{2 }- ac = 0.

Question: 10

If p, q are the real roots and p = q. Then show that the roots of the equation (p-q)x^{2 }+ 5(p + q)x - 2(p - q) = 0 are real and equal.

Solution:The given equation is (p - q)x

^{2}+ 5(p + q)x - 2(p - q) = 0Given, p , q are real and p ? q.

Then, Discriminant (D) = b

^{2}– ac = [5(p + q)]^{2}- 4(p - q)(-2(p - q)) = 25(p + q)^{2}+ (p - q)^{2}We know that the square of any integer is always positive that is, greater than zero.

Hence, (D) = b

^{2 }–ac = 0 As given, p, q are real and p = q.Therefore, = 25(p + q)

^{2}+ (p - q)^{2 }? 0 = D = 0Therefore, the roots of this equation are real and unequal.

Question: 11

If the roots of the equation (c^{2 }- ab)x^{2 }- 2(a^{2 }- bc)x + b^{2 }- ac = 0 are equal , then prove that either a = 0 or a³ + b³ + c³ = 3abc .

Solution:The given equation is (c

^{2 }- ab)x^{2 }- 2(a^{2 }- bc)x + b^{2 }- ac = 0This equation is in the form of ax

^{2 }+ bx + c = 0So, a = (c

^{2 }- ab), b = -2(a^{2 }- bc), c = b^{2 }- ac.According to the question, the roots of the given question are equal.

Hence, D= 0, b

^{2 }- 4ac = 0= [-2(a

^{2 }- bc)]^{2}- 4(c^{2 }- ab)( b^{2 }- ac) = 0= 4(a

^{2 }- bc)^{2}- 4(c^{2 }- ab)( b^{2 }- ac) = 0= 4a(a³ + b³ + c³ - 3abc) = 0

Either 4a =0 therefore, a = 0 Or, (a³ + b³ + c³ - 3abc) = 0

= (a³ + b³ + c³) = 3abc Hence its is proved.

Question: 12Show that the equation 2(a

^{2 }+ b^{2})x^{2 }+ 2(a + b)x + 1 = 0 has no real roots , when a = b.

Solution:The given equation is 2(a

^{2 }+ b^{2})x^{2 }+ 2(a + b)x - 1 = 0This equation is in the form of ax

^{2 }+ bx + c = 0Here, a = 2(a

^{2 }+ b^{2}), b = 2(a + b), c = +1.Given, a = b The discriminant (D) = b

^{2 }- 4ac = [2(a + b)]^{2}- 4(2(a^{2 }+ b^{2}))(1)= 4(a + b)

^{2}- 8(a^{2 }+ b^{2})= 4(a

^{2 }+ b^{2 }+ 2ab) – 8a^{2 }- 8b^{2}= +2ab – 4a

^{2 }- 4b^{2}According to the question a = b, as the discriminant D has negative squares so the value of D will be less than zero. Hence, D = 0, when a = b.

Question: 13

Prove that both of the roots of the equation (x - a)(x - b) +(x - c)(x - b) + (x - c)(x - a) = 0 are real but they are equal only when a = b = c.

Solution:The given equation is (x - a)(x - b) + (x - c)(x - b) + (x - c)(x - a) = 0

By solving the equation, we get it as, 3x

^{2 }- 2x(a + b + c) + (ab + bc + ca) = 0This equation is in the form of ax

^{2 }+ bx + c = 0Here, a = 3, b = 2(a + b + c), c = (ab + bc + ca)

The discriminate (D) = b

^{2 }- 4ac= [-2(a + b + c)]

^{2 }- 4(3)(ab + bc + ca)= 4(a + b +c )

^{2 }-12(ab + bc + ca)= 4[(a + b + c)

^{2 }- 3(ab + bc + ca)]= 4[a

^{2 }+ b^{2 }+ c^{2 }– ab - bc - ca]= 2[2a

^{2 }+ 2b^{2 }+ 2c^{2 }– 2ab - 2bc - 2ca]= 2[(a - b)

^{2 }+ (b - c)^{2 }+ (c - a)^{2}]Here clearly D = 0, if D = 0 then, [(a - b)

^{2 }+ (b - c)^{2 }+ (c - a)^{2}] = 0a –b = 0 b – c = 0 c – a = 0

Hence, a = b = c = 0 Hence, it is proved.

Question: 14

If a, b, c are real numbers such that ac = 0, then, show that at least one of the equations ax^{2 }+ bx + c = 0 and – ax^{2 }+ bx + c = 0 has real roots.

Solution:The given equation are ax

^{2 }+ bx + c = 0 … (i)And - ax

^{2 }+ bx + c = 0 … (ii)Given, equations are in the form of ax

^{2}+ bx + c = 0 also given that a, b, c are real numbers and ac = 0.The Discriminant (D) = b

^{2 }- 4ac For equation (i) = b^{2 }- 4ac ... (iii)For equation (ii) = b

^{2 }- 4(-a)(c) = b^{2 }+ 4ac … (iv)As a, b, c are real and given that ac = 0

Hence b

^{2 }- 4ac = 0 and b^{2}+ 4ac = 0 Therefore, D = 0 Hence proved.

Question: 15If the equation (1 + m

^{2})x + 2mcx + (c^{2 }- a^{2}) = 0 has real and equal roots , prove that c^{2 }= a^{2}(1 + m^{2}).

Solution:The given equation is (1 + m

^{2})x^{2 }+ 2mcx + (c^{2 }- a^{2}) = 0The above equation is in the form of ax

^{2 }+ bx + c = 0.Here a = (1 + m

^{2}), b = 2mc, c = +(c^{2 }- a^{2})Given, that the nature of the roots of this equation is equal and hence D = 0, b

^{2 }- 4ac = 0= (2mc)

^{2}- 4(1 + m^{2})(c^{2 }- a^{2}) = 0= 4m

^{2}c^{2 }– 4(c^{2 }+ m^{2}c^{2 }- a^{2}– a^{2}m^{2}) = 0= 4(m

^{2}c^{2 }- c^{2 }+ m^{2}c^{2 }+ a^{2}+ a^{2}m^{2}) = 0= m

^{2}c^{2 }- c^{2 }+ m^{2}c^{2 }+ a^{2}+ a^{2}m^{2 }= 0Now cancelling out the equal and opposite terms, = a

^{2}+ a^{2}m^{2 }- c^{2 }= 0= a

^{2}(1 + m^{2}) – c^{2}= 0Therefore, c

^{2 }= a^{2}(1 + m^{2}) Hence it is proved that as D = 0, then the roots are equal of c^{2 }= a^{2}(1+ m^{2}).