How do you solve $16 x^{2} - 81 \geq 0$ $16x^2-81>=0$ ?

Question

How do you solve $16 x^{2} - 81 \geq 0$ $16x^2-81>=0$ ?

1 Answer

Impact of this question

1436 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-01T15:08:42

The solution is $x \in] - \infty, - \frac{9}{4}] \cup [\frac{9}{4}, + \infty [$ $x in ]-oo,-9/4]uu[9/4,+oo[$

Explanation:

We need

$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

Let's factorise the inequality

$16 x^{2} - 81 = (4 x + 9) (4 x - 9)$ $16x^2-81=(4x+9)(4x-9)$

Let $f (x) = (4 x + 9) (4 x - 9)$ $f(x)=(4x+9)(4x-9)$

We build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- \frac{9}{4}$ $-9/4$ $a a a a$ $color(white)(aaaa)$ $\frac{9}{4}$ $9/4$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $4 x + 9$ $4x+9$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $4 x - 9$ $4x-9$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

Therefore,

$f (x) \geq 0$ $f(x)>=0$ when $x \in] - \infty, - \frac{9}{4}] \cup [\frac{9}{4}, + \infty [$ $x in ]-oo,-9/4]uu[9/4,+oo[$

Science

Math

Humanities

... and beyond

How do you solve $16 x^{2} - 81 \geq 0$ $16x^2-81>=0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 16x^2-81>=016x2−81≥016x^2-81>=0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $16 x^{2} - 81 \geq 0$ $16x^2-81>=0$ ?