Question

Find the value of k for which the equation X square + 5kx + 16 is equal to zero has real and equal root

Answer 1

b2-4ac=0

25k2-64=0

5k whole squre=64

$y = \sqrt{64 \div 25}$

y=k

$y = + - 8 \div 5$

Answer 2

${x}^{2} + 5kx + 16 \\ a {x}^{2} + bx + c \\ a = 1 \: \: \: b = 5k \: \: \: \: c = 16$

we know that if

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

then the roots are real and equal.

${b}^{2} - 4ac = 0 \\ {(5k)}^{2} - 4(1)(16) = 0 \\ 25 {k}^{2} - 64 = 0 \\ 25 {k}^{2} = 64 \\ {k}^{2} = \frac{64}{25} \\ k = + or - \frac{8}{5}$