Question

the roots of X square +kx+k = 0 are real and equal, find k.

Answer 1

If the equation has real and equal roots, then
D = 0
=> b² - 4ac = 0
a=1 , b = k and c =k
(k)²-4(1)(k)=0
k²-4k=0
k(k-4) =0
k=0
and
k-4=0
k=4

Answer 2

Answer:

k = 0 Or k = 4

Explanation:

Given quadratic equation :

x²+kx+k=0

Compare above equation with

ax²+bx+c=0 , we get

a = 1 , b = k , c = k

Now ,

Discreminant (D) = 0

[ given roots are real and equal ]

=> b²-4ac = 0

=> k²-4×1×k = 0

=> k(k-4)=0

=> k = 0 Or k-4 = 0

=> k = 0 Or k = 4

Therefore,.

k = 0 Or k = 4

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