How do you solve $9 x^{2} + 3 x - 2 \geq 0$ $9x^2+3x-2>=0$ using a sign chart?

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How do you solve $9 x^{2} + 3 x - 2 \geq 0$ $9x^2+3x-2>=0$ using a sign chart?

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Answer 1 · 2017-01-13T16:19:20

The answer is $x \in] - \infty, - \frac{2}{3}] \cup [\frac{1}{3}, + \infty [$ $x in ] -oo,-2/3 ] uu [1/3, +oo[$

Explanation:

Let's factorise the LHS

$9 x^{2} + 3 x - 2 = (3 x - 1) (3 x + 2)$ $9x^2+3x-2=(3x-1)(3x+2)$

Let $f (x) = (3 x - 1) (3 x + 2)$ $f(x)=(3x-1)(3x+2)$

Now we can make the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a a$ $color(white)(aaaaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- \frac{2}{3}$ $-2/3$ $a a a a a a$ $color(white)(aaaaaa)$ $\frac{1}{3}$ $1/3$ $a a a a a$ $color(white)(aaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $3 x + 2$ $3x+2$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $3 x - 1$ $3x-1$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $+$ $+$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$

Therefore,

$f (x) \geq 0$ $f(x)>=0$ , when $x \in] - \infty, - \frac{2}{3}] \cup [\frac{1}{3}, + \infty [$ $x in ] -oo,-2/3 ] uu [1/3, +oo[$

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How do you solve $9 x^{2} + 3 x - 2 \geq 0$ $9x^2+3x-2>=0$ using a sign chart?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve 9x2+3x−2≥09x^2+3x-2>=0 using a sign chart?

1 Answer

Explanation:

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Impact of this question

How do you solve $9 x^{2} + 3 x - 2 \geq 0$ $9x^2+3x-2>=0$ using a sign chart?