Question

Find the value of 'k' for which the equation x2-4x+k=0 has equal roots.

Answer 1

D = 0
b^2 -4ac =0
(-4)^2 -4(1)(k) =0
16-4k=0
16=4k
k=4


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Answer 2

Answer:

$The \: value \: of \: k = 4$

Step-by-step explanation:

$Given\\ \: quadratic \: equation \: x^{2}-4x+k=0$

\* Compare above equation with ax²+bx+c=0, we get

a = 1 , b = -4 , c = k

$Discreminant (D)=0$

/* Given roots are equal */

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

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

$\implies 16-4k=0$

$\implies -4k = -16$

$\implies k = \frac{-16}{-4}$

$\implies k = 4$

Therefore,.

$The \: value \: of \: k = 4$

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