Question

 If the roots of the quadratic equation 9x² + kx + k = 8 are real and equal, what is the value of k?​

Answer 1

9x² + kx + k = 8

=> 9x² + kx + k-8=0

Here , a = 9 , b = k and c = k-8

For real and equal roots :

• D = 0

=> b² - 4 ac = 0

=> k² - 4 (9)(k-8)= 0

=> k² -36k + 288 = 0

=> k² -24k - 12k +288=0

=> k ( k-24 ) -12 (k- 24) =0

=> (k-24)(k-12)=0

=> k = 12 and 24

Answer 2

Given :-

If the roots of the quadratic equation 9x² + kx + k = 8 are real and equal,

To Find :-

what is the value of k?​

Solution :-

We know that

9x² + kx + k = 8

9x² + kx + k - 8 = 0

D = b² - 4ac

Here

b = k

a = 9

c = k - 8

By putting values

0 = (k)² - 4(9)(k - 8)

0 = k² - 36(k - 8)

0 = k² - 36k + 288

By quadratic formula

x = -b ± √(b² - 4ac)/2a

x = -(-36) ± √(36² - 4(1)(288))/2(1)

x = 36 ± √(1296 - 1152)/2

x = 36 ± √(144)/2

x = 36 ± 12/2

x = 36 + 12/2 or 36 - 12/2

x = 48/2 or 24/2

x = 24 or 12