Question

what is the value of k, if the roots of x2+kx+k=0 are real and equal
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Answer 1

Answer:

k = 0 or k = 4

Step-by-step explanation:

Given $x^{2} +kx+k$

roots are real & equal

$bx^{2} - 4ac = 0$

$=(k)^{2}- 4 (1)(k)=0$

$=k^{2} - 4k=0$

$=k(k-4)=0$

∴  k = 0 or k = 4

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Answer 2

Answer:

0 or 4

Step-by-step explanation:

x²+Kx+k has a distinct real solution.

For a distinct real solution we use the formula b²-4ac>0

Here a=1,b=k and c= k

By solving this we get

k²-4k>0

k(k-4)>0

k>0 , k<4.