Question

find the value of k for which equation 4x^2+8x-k=0 has real roots​

Answer 1

Answer:

Hope this was helpful

Answer 2

Given equation,

$4{x}^{2} + 8x - k = 0$

a = 4

b = 8

c = -k

Roots are real.

So,

$\boxed {{b}^{2} - 4ac \geq 0} \\ \implies {8}^{2} - 4(4)(-k) \geq 0 \\ \implies 64 + 16k \geq 0 \\ \implies k \leq \dfrac{-64}{16} \leq -4$

Value of $k \leq -4$