Manipulating inequalities with square roots

Question

A curve has the equation $$y = 3x^2 -2,$$ and a straight line has the equation: $$y = mx - 5 .$$

These lines do not meet, and to find the values of $m$, I am aware that you could find the values they would meet at, and then do a counterargument to say they wouldn't meet at all other values.

I was able to do $3x^2 - 2 = mx - 5$, when they meet to create $$3x^2 -mx + 3 = 0,$$ and then I used $b^2 - 4ac < 0$, to say that when the discriminant is less than $0$, there will be no roots and that is when the curve and line do not meet.

No, my problem is so trivial: When you fill the values in you get $$m^2 - 36 < 0.$$ Square rooting this gives $m < 6$ and $m < -6$.

The mark scheme says the answer is $-6 < m < 6$. Why are the signs arranged this way? How do I get from what I get to that?

Answer 1

There is just one mistake. You cannot square root both sides of $m^2<36$ to get $m<6$ and $m<-6$. Instead, factor $m^2-36$: $$(m+6)(m-6)<0.$$ If the product of two factors is negative, then one factor is positive and the other is negative. If $m+6 > 0$ and $m-6 < 0$, then we must have $-6 < m < 6$. We should also check the other possibility, i.e., $m+6 < 0$ and $m-6 > 0$, but this yields no solutions. So the desired range is $-6 < m < 6$.

Answer 2

Rearranging the equation $m^2 - 36 < 0$ gives $m^2 < 36$. Since $\sqrt{\cdot}$ is an increasing function, $$|m| = \sqrt{m^2} < \sqrt{36} = 6 ,$$ and $|m| < 6$ is equivalent to $$-6 < m < 6 .$$

Answer 3

When solving a quadratic inequality it is helpful to sketch a graph of the function, in this case $y=m^2-36=(m+6)(m-6)$ and observe the values of $m$ for which, in this case, $y<0$. Since it is a U-shaped graph, the solution set is $-6<m<6$