What is (4x^2-4x-8)/(-4x+8)<= 0 ?

The solution to (4x^2-4x-8)/(-4x+8) =-1

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(4x^2-4x-8)/(-4x+8)<= 0

\frac{4 x ^{2} - 4 x - 8}{- 4 x + 8} \leq 0

x \geq - 1

Interval Notation

[- 1, \infty)

Solution steps

\frac{4 x ^{2} - 4 x - 8}{- 4 x + 8} \leq 0

Factor

\frac{4 x ^{2} - 4 x - 8}{- 4 x + 8} : \frac{4 ( x + 1 ) ( x - 2 )}{- 4 ( x - 2 )}

\frac{4 ( x + 1 ) ( x - 2 )}{- 4 ( x - 2 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{4 ( x + 1 ) ( x - 2 ) ( - 1 )}{- 4 ( x - 2 )} \geq 0 \cdot (- 1)

Simplify

\frac{4 ( x + 1 ) ( x - 2 ) ( - 1 )}{- 4 ( x - 2 )} : x + 1

x + 1 \geq 0

Identify the intervals

x \geq - 1

4 x^{2} - 8 x + 63 = 0

2 a^{2} - 3 a - 2

x^{2} = 27

- 3 (4 b + 1) = - 51

- 4 - 4

What is (4x^2-4x-8)/(-4x+8)<= 0 ?
The solution to (4x^2-4x-8)/(-4x+8)<= 0 is x>=-1