Question

Find the values of k for which the equation kx 2-kx+1=0

Answer 1

Answer:

$k = 4$

Step-by-step explanation:

To find the values of k for which the equation

$k {x}^{2} - kx + 1 = 0$

has real and equal roots.

We know that a Quadratic equation has equal roots if D= 0

$D =0 \\ \\ {b}^{2} - 4ac = 0 \\ \\k {x}^{2} - kx + 1 \\ \\ a = k \\ \\ b = - k \\ \\ c = 1 \\ \\ {( - k)}^{2} - 4(k)(1) = 0 \\ \\ {k}^{2} - 4k = 0 \\ \\ k(k - 4) = 0 \\ \\ k = 0 \\ \\ or \\ \\ k = 4$

Here we have to discard k=0,because then the equation will not be a Quadratic equation.

So,

$\boxed{k = 4}$

is the right answer.

Hope it helps you.

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**find the value of k for which the quadratic equation is

$(k+4)x^2+(k+1)x+1=0$

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