Question

Find the value for which roots of q.e kx^2-5x+k=0 are real and equal

Answer 1

discriminant = 0

25 - 4 k.k= 0

k = 5/2

Answer 2

For real and equal roots discriminant is zero.
i.e
${b}^{2} - 4ac = 0 \\ = > {( - 5)}^{2} - 4 \times k \times k = 0 \\ = > 25 = 4 \times {k}^{2} \\ = > \sqrt{ \frac{25}{4} } = k \\ = > \frac{5}{2} = k$
therefore, k = 5/2