FIND THE VALUE OF K IN THE QUADRATIC EQUATION

Find values of k such that :

Example 1 :

kx² + 12x + 2  =  0 has a repeated root

Solution :

Given, kx² + 12x + 2  =  0

By comparing ax² + bx + c  =  0 in the given equation, we get

Here, a  =  k, b  =  12, and c  =  2

Since the roots are real and equal, then D  =  0

b² – 4ac  =  0

(12)² – 4(k)(2)  =  0

144 – 8k  =  0

-8k  =  -144

k  =  144/8

k  =  18

So, the value of k is 18

Example 2 :

3x² + 16x + k  =  0 has a repeated root

Solution :

Given, 3x² + 16x + k  =  0

By comparing ax² + bx + c  =  0 in the given equation, we get

Here, a  =  3, b  =  16, and c  =  k

Since the roots are real and equal, then D  =  0

b² – 4ac  =  0

(16)² – 4(3)(k)  =  0

256 – 12k  =  0

-12k  =  -256

k  =  (256)/12

k  =  64/3

By writing a mixed fraction, we get

k  =  21 1/3

So, the value of k is 21 1/3

Example 3 :

kx² + 6x - 3  =  0 has two distinct real roots

Solution :

Given, kx² + 6x - 3  =  0

By comparing ax² + bx + c  =  0 in the given equation, we get

Here, a  =  k, b  =  6, and c  =  -3

Since the roots are real and distinct, then D > 0

b² – 4ac > 0

(6)² – 4(k)(-3) > 0

36 + 12k > 0

12k > -36

k > -36/12

k > -3

So, the solution of k is k > -3, k ≠ 0

Example 4 :

4x² - 12x + k  =  0 has two distinct real roots

Solution :

Given, 4x² - 12x + k  =  0

By comparing ax² + bx + c  =  0 in the given equation, we get

Here, a  =  4, b  =  -12, and c  =  k

Since the roots are real and equal, then D > 0

b² – 4ac > 0

(-12)² – 4(4)(k) > 0

144 – 16k > 0

-16k > -144

k < (144)/16

k < 9

So, the solution of k is k < 9

Example 5 :

2x² - 5x + k  =  0 has no real solutions

Solution :

Given, 2x² - 5x + k  =  0

By comparing ax² + bx + c  =  0 in the given equation, we get

Here, a  =  2, b  =  -5, and c  =  k

Since the roots are not real, then D < 0

b² – 4ac < 0

(-5)² – 4(2)(k) < 0

25 – 8k < 0

-8k < -25

k > 25/8

By writing a mixed fraction, we get

k > 3 1/8

So, the solution of k is k > 3 1/8

Example 6 :

kx² - 11x - k  =  0 has no real roots.

Solution :

Given, kx² - 11x - k  =  0

By comparing ax² + bx + c  =  0 in the given equation, we get

Here, a  =  k, b  =  -11, and c  =  -k

Since the roots are not real, then D < 0

b² – 4ac < 0

(-11)² – 4(k)(-k) < 0

121 + 4k² < 0

4k² < -121

k² < -(121/4)

k < √(-121/4)

Since we have a root that is imaginary, so there is no solution.

Example 7 :

The value of k  for which x = -2 is a root of the quadratic equation kx² + x - 6 = 0.

Solution :

kx² + x - 6 = 0

Since -2 is a root of the polynomial, then 

k(-2)² + (-2) - 6 = 0

4k - 2 - 6 = 0

4k - 8 = 0

4k = 8

k = 8/4

k = 2

So, the required value of k is 2.

Example 8 :

px² + 3x + q = 0 has two roots x = -1 and x = -2, then the value of p - q is.

Solution :

Given quadratic equation is,

px² + 3x + q = 0

When x = -1

p(-1)² + 3(-1) + q = 0

p - 3 + q = 0

p + q = 3 ------(1)

When x = -2

p(-2)² + 3(-2) + q = 0

4p - 6 + q = 0

4p + q = 6 ------(2)

(1) - (2)

p + q - 4p - q = 3 - 6

-3p = -3

p = 1

Applying the value of p, we get

1 + q = 3

q = 3 - 1

q = 2

p - q = 1 - 2

= -1

So, the answer is -1.

Example 9 :

The value of p so that the quadratic equation 

x² + 5px + 16 = 0

has no real root is

a)  p > 8     b)  p < 5     c)  -8/5 < p < 8/5    d) -8/5 ≤ x < 0

Solution :

x² + 5px + 16 = 0

Since the quadratic equation has no real roots, then 

b² - 4ac < 0

a = 1, b = 5p and c = 16

(5p)² - 4(1)(16) < 0

25p² - 64 < 0

25p² < 64

p² < 64/25

p < ±8/5

-8/5 < p < 8/5

So, option c is correct.

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