How do you solve $4 t^{2} > 12$ $4t ^ { 2} > 12$ ?

Question

How do you solve $4 t^{2} > 12$ $4t ^ { 2} > 12$ ?

2 Answers

Impact of this question

1521 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-10T13:38:33

$t > \sqrt{3}$ $t>sqrt3$ OR $t < - \sqrt{3}$ $t<-sqrt3$

Explanation:

We have the inequality:

$4 t^{2} > 12$ $4t^2>12$

1) Divide both sides by $4$ $4$ :

$\frac{4 t^{2}}{4} > \frac{12}{4}$ $(4t^2)/4>12/4$

$\Rightarrow t^{2} > 3$ $=>t^2>3$

2) Take the square root of both sides:

$\sqrt{t^{2}} > \sqrt{3}$ $sqrt(t^2)>sqrt(3)$

$\Rightarrow t > \sqrt{3}$ $=>t>sqrt3$

BUT square roots have two solutions - positive or negative. You know that, when we divide by a negative, we change the sign of the inequality. So:

$t > \sqrt{3}$ $t>sqrt3$ OR $t < - \sqrt{3}$ $t<-sqrt3$

Answer 2 · 2018-03-10T14:13:32

The solution is $t \in (- \infty, - \sqrt{3}) \cup (\sqrt{3}, + \infty)$ $t in (-oo, -sqrt3) uu(sqrt3, +oo)$

Explanation:

Let's rearrange the inequality

$4 t^{2} > 12$ $4t^2>12$

$4 t^{2} - 12 > 0$ $4t^2-12>0$

Factorise,

$4 (t^{2} - 3) > 0$ $4(t^2-3)>0$

$4 (t + \sqrt{3}) (t - \sqrt{3}) > 0$ $4(t+sqrt3)(t-sqrt3)>0$

Let $f (t) = 4 (t + \sqrt{3}) (t - \sqrt{3})$ $f(t)=4(t+sqrt3)(t-sqrt3)$

Now build a sign chart

$a a a a$ $color(white)(aaaa)$ $t$ $t$ $a a a a a a$ $color(white)(aaaaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- \sqrt{3}$ $-sqrt3$ $a a a a$ $color(white)(aaaa)$ $\sqrt{3}$ $sqrt3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $t + \sqrt{3}$ $t+sqrt3$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $t - \sqrt{3}$ $t-sqrt3$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (t)$ $f(t)$ $a a a a a a a a a a$ $color(white)(aaaaaaaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (t) > 0$ $f(t)>0$ when $t \in (- \infty, - \sqrt{3}) \cup (\sqrt{3}, + \infty)$ $t in (-oo, -sqrt3) uu(sqrt3, +oo)$

graph{4x^2-12 [-25.65, 25.66, -12.83, 12.84]}

Science

Math

Humanities

... and beyond

How do you solve $4 t^{2} > 12$ $4t ^ { 2} > 12$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 4t2>124t ^ { 2} > 12?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve $4 t^{2} > 12$ $4t ^ { 2} > 12$ ?