Question

What is the solution of x^2-1/x^2+5x+4< or equal to 0

Answer 1

there is not a solution

Answer 2

Answer:

(-4,-1) U (-1,1]

Step-by-step explanation:

x^2-1/x^2+5x+4< or equal to 0

$\frac{x^2-1}{x^2+5x+4} \leq 0$

LEts factor top and bottom

$x^2+5x+4= (x+4)(x+1)$

Now we factor x^2-1 using $a^2-b^2=(a+b)(a-b)$

$x^2-1^2=(x+1)(x-1)$

now we find the x values that makes the denominator 0

$(x+4)(x+1)=0$

x+4=0, x=-4

x+1=0, x=-1

$\frac{(x+1)(x-1) }{(x+4)(x+1)} =0$

multiply the denominator on both sides

cancel out x+1 at the top and bottom

x-1 =0, x=1

We got 3 x values

x=-4, -1, 1

using the x values we make 4 intervals

(-∞, -4), (-4,-1) (-1,1] and [1,∞)

LEts pick a random number from each interval and check with the inequality

(-∞, -4) pick -5

$\frac{(-5)^2-1}{(-5)^2+5(-5)+4} \leq 0$

$\frac{24}{4} \leq 0$  false

(-4,-1) pick -3

$\frac{(-3)^2-1}{(-3)^2+5(-3)+4} \leq 0$

$\frac{8}{-1} \leq 0$  True

(-1,1] pick 0

$\frac{(0)^2-1}{(0)^2+5(0)+4} \leq 0$

$\frac{-1}{4} \leq 0$  True

[1,∞) pick -2

$\frac{(2)^2-1}{(2)^2+5(2)+4} \leq 0$

$\frac{3}{18} \leq 0$  false

solution are the intervals that satisfies our inequality

(-4,-1) U (-1,1]