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)##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]}