the roots of X square +kx+k = 0 are real and equal, find k.
Answers
Answered by
284
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
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
Answered by
120
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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