How do you solve $3 (4 - x) (2 x + 1) < 0$ $3(4 - x)(2x + 1)<0$ ?

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How do you solve $3 (4 - x) (2 x + 1) < 0$ $3(4 - x)(2x + 1)<0$ ?

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Answer 1 · 2016-03-14T17:16:01

x > 4 and x < $- \frac{1}{2}$ $-1/2$ . In other words, x is outside $[- \frac{1}{2}, 4]$ $[ -1/2, 4 ]$

Explanation:

Drop the positive factor 3.
$- 2 x^{2} + 7 x + 4 < 0$ $-2 x^2 + 7 x +4 < 0$ .
${(x - \frac{7}{4})}^{2} > \frac{81}{16}$ $( x - 7/4 )^2 > 81/16$
$∣ ∣ ∣ x - \frac{7}{4} ∣ ∣ ∣ > \frac{9}{4}$ $| x - 7/4 | > 9/4$ .
If $| x - a | > b, x > a + b and x < a - b$ $| x - a | > b, x > a + b and x < a - b$
Here, $x > 4 and x < - \frac{1}{2}$ $x > 4 and x < -1/2$ .

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How do you solve $3 (4 - x) (2 x + 1) < 0$ $3(4 - x)(2x + 1)<0$ ?

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How do you solve 3(4−x)(2x+1)<0 3(4 - x)(2x + 1)<0?

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Explanation:

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How do you solve $3 (4 - x) (2 x + 1) < 0$ $3(4 - x)(2x + 1)<0$ ?