Question

The values of k for which the quadratic equation 16x²+4kx+9=0 has real and equal roots are
(a)6, $\frac{-1}{6}$
(b)36, −36
(c)6, −6
(d)34, -$\frac{-1}{6}$

Answer 1

SOLUTION :  

Option (c) is correct : 6 , - 6

Given : 16x² + 4kx + 9 = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 16 , b = 4k , c = 9

D(discriminant) = b² – 4ac

D = (4k)² - 4 × 16 × 9

D = 16k² - 576

Given : Roots are real and equal i.e D = 0.

16k² - 576 = 0

16(k² - 36) = 0

k² - 36 = 0

(k)² - (6)² = 0

(k - 6) (k + 6) = 0

[(a² - b²) = (a + b)(a - b)]

k - 6 = 0  or  k + 6 = 0

k = 6  or  k = - 6

Hence, the value of k is 6 & - 6 .

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Solution :

Compare given Quadratic

equation 16x²+4kx+9=0

with ax²+bx+c = 0 , we get

a = 16 , b = 4k , c = 9

Discreminant (D)=0

[ Given ,roots are real and

equal ]

b² - 4ac= 0

=> (4k)² - 4×16×9 = 0

=> 16k² - 16×4×9 = 0

=> 16( k² - 36 ) = 0

=> k² - 36 = 0

=> k² = 36

=>k = ± √36

=> k = ± 6