How do you solve the inequality $5 x + 4 \leq 3 x^{2}$ $5x+4<=3x^2$ and write your answer in interval notation?

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How do you solve the inequality $5 x + 4 \leq 3 x^{2}$ $5x+4<=3x^2$ and write your answer in interval notation?

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Answer 1 · 2017-12-11T10:52:02

The solution is $x \in (- \infty, - 0.59] \cup [2.26, + \infty)$ $x in (-oo, -0.59] uu[2.26, +oo)$

Explanation:

Let's rewrite the inequality

$5 x + 4 \leq 3 x^{2}$ $5x+4<=3x^2$

$3 x^{2} - 5 x - 4 \geq 0$ $3x^2-5x-4>=0$

Let $f (x) = 3 x^{2} - 5 x - 4$ $f(x)=3x^2-5x-4$

The roots of the quadratic equation $3 x^{2} - 5 x - 4 = 0$ $3x^2-5x-4=0$ , are

$x = \frac{5 \pm \sqrt{{(- 5)}^{2} - (4) \cdot (3) \cdot (- 4)}}{6} = \frac{5 \pm \sqrt{73}}{6}$ $x = ( 5+-sqrt ((-5)^2-(4)* (3) * (-4)) ) /(6)=(5+-sqrt(73))/(6)$

$x_{1} = \frac{5 - \sqrt{73}}{6} = - 0.59$ $x_1=(5-sqrt73)/6=-0.59$

$x_{2} = \frac{5 + \sqrt{73}}{6} = 2.26$ $x_2=(5+sqrt73)/6=2.26$

Let's build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a$ $color(white)(aaaaa)$ $- \infty$ $-oo$ $a a a a a a a$ $color(white)(aaaaaaa)$ $x_{1}$ $x_1$ $a a a a a a$ $color(white)(aaaaaa)$ $x_{2}$ $x_2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x - x_{1}$ $x-x_1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $0$ $0$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - x_{2}$ $x-x_2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ #color(white)(aaaa)-# $a a$ $color(white)(aa)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $0$ $0$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a$ $color(white)(aa)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$

Therefore,

$f (x) \geq 0$ $f(x)>=0$ when $x \in (- \infty, x_{1}] \cup [x_{2}, + \infty)$ $x in (-oo, x_1] uu[x_2, +oo)$

graph{3x^2-5x-4 [-11.39, 11.11, -6.615, 4.635]}

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How do you solve the inequality $5 x + 4 \leq 3 x^{2}$ $5x+4<=3x^2$ and write your answer in interval notation?

1 Answer

Explanation:

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How do you solve the inequality $5 x + 4 \leq 3 x^{2}$ $5x+4<=3x^2$ and write your answer in interval notation?