Question

find the value of k for which the equation X square minus kx + 1 is equal to zero has real and equal roots ​

Answer 1

Answer:

$k = 2$

Step-by-step explanation:

If roots are real and equal, then the condition is

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

So, in given equation

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