Question

The positive value of k for which x² + kx + 64 = ,0 & x² - 8x + k = 0 will have real roots. ​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {x^2 + Kx + 64 = 0}$

$\implies \mathsf {b^2 - 4ac \geqslant 0}$

$\mathsf {K^2 - 256 \geqslant 0}$

$\mathsf {K \geqslant 16 \:or\:K \leqslant - 16..........(1)}$

$\mathsf {x^2 - 8x + K = 0}$

$\mathsf {64 - 4K \geqslant 0}$

$\implies \mathsf {4K \leqslant 64}$

$\mathsf {K \leqslant 16..........(2)}$

$\mathsf \blue{From\:(1)\:and \:(2)\:K = 16}$

Answer 2

as we know,if there are two equations a1x+b1y+c1=0 and ax2+by2+c2=0

if there roots are real and equal then

a1/a2 =b1/b2=c1/c2

so

here

1/1=k/-8

so k=-8

hope it helps you