Question

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

Answer 1

`t>sqrt3` OR `t<-sqrt3`

Explanation:

We have the inequality:

`4t^2>12`

1) Divide both sides by `4`:

`(4t^2)/4>12/4`

`=>t^2>3`

2) Take the square root of both sides:

`sqrt(t^2)>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>sqrt3` OR `t<-sqrt3`

Answer 2

The solution is `t in (-oo, -sqrt3) uu(sqrt3, +oo)`

Explanation:

Let's rearrange the inequality

`4t^2>12`

`4t^2-12>0`

Factorise,

`4(t^2-3)>0`

`4(t+sqrt3)(t-sqrt3)>0`

Let `f(t)=4(t+sqrt3)(t-sqrt3)`

Now build a sign chart

`color(white)(aaaa)` `t` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `-sqrt3` `color(white)(aaaa)` `sqrt3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `t+sqrt3` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `t-sqrt3` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(t)` `color(white)(aaaaaaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(t)>0` when `t in (-oo, -sqrt3) uu(sqrt3, +oo)`

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