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

2 Answers
Mar 10, 2018

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

Mar 10, 2018

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)tcolor(white)(aaaaaa)-oocolor(white)(aaaa)-sqrt3color(white)(aaaa)sqrt3color(white)(aaaa)+oo

color(white)(aaaa)t+sqrt3color(white)(aaaaaaa)-color(white)(aaaa)+color(white)(aaaa)+

color(white)(aaaa)t-sqrt3color(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]}