How do you solve $6 x^{2} - 5 x - 4 \leq 0$ $6x^2-5x-4<=0$ ?

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How do you solve $6 x^{2} - 5 x - 4 \leq 0$ $6x^2-5x-4<=0$ ?

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Answer 1 · 2015-05-05T17:20:37

It is fairly easy to factor the left side of this inequality
$6 x^{2} - 5 x - 4 \leq 0$ $6x^2 -5x-4<=0$
becomes
$(3 x - 4) (2 x + 1) \leq 0$ $(3x-4)(2x+1) <= 0$

The trick is in seeing what this means.

For this to be true
either $(3 x - 4) \leq 0 \to x \leq \frac{4}{3}$ $(3x-4)<=0 rarr x<=4/3$
or $(2 x + 1) \leq 0 \to x < - \frac{1}{2}$ $(2x+1)<=0 rarr x<-1/2$
BUT NOT both $< 0$ $<0$

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How do you solve $6 x^{2} - 5 x - 4 \leq 0$ $6x^2-5x-4<=0$ ?

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