How do you solve $x^{2} + 5 x + 4 \leq 0$ $x^2+5x+4<=0$ ?

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How do you solve $x^{2} + 5 x + 4 \leq 0$ $x^2+5x+4<=0$ ?

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Answer 1 · 2016-06-26T19:53:49

$x \in [- 4, - 1]$ $x in [-4, -1]$

Explanation:

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

This is zero when $x = - 4$ $x=-4$ or $x = - 1$ $x=-1$

The quadratic has a positive leading coefficient, so is is an upright parabola, crossing the $x$ $x$ axis at $(- 4, 0)$ $(-4, 0)$ and $(- 1, 0)$ $(-1, 0)$ .

Hence any $x \in [- 4, - 1]$ $x in [-4, -1]$ satisfies the required inequality and any value of $x$ $x$ outside that closed interval does not.

graph{x^2+5x+4 [-10, 10, -5, 5]}

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How do you solve $x^{2} + 5 x + 4 \leq 0$ $x^2+5x+4<=0$ ?

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How do you solve x2+5x+4≤0x^2+5x+4<=0?

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How do you solve $x^{2} + 5 x + 4 \leq 0$ $x^2+5x+4<=0$ ?