Question

the roots of the given quadratic equation is real and equal find k for y square + k y + 25 = 20​

Answer 1

$y {}^{2} + ky + 25 = 20 \\ y {}^{2} + ky + (25 - 20) = 0 \\ \\ now \: \: \: a = 1 \: \: b = k\: \: c = 5 \\ b {}^{2} - 4ac = 0 \: \: \: \: \: (given) \\ (k {}^{2} ) - 4(1)(5) = 0 \\ k {}^{2} - 20 = 0 \\ k {}^{2} = 20 \\ k = 2 \sqrt{5}$