Question

roots of x square + kx + k =0 are real and equal, find k​

Answer 1

Answer:

for equal roots the condition is

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

So the value of k is

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

$\huge{\mathcal{\blue{\boxed{k=4}}}}$

Answer 2

Answer:

$\sqrt{ {x}^{2} } + kx + k = 0$

Step-by-step explanation:

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