How do I solve the inequality $2(x-3)^2 < 32$?

Question

How do I solve this inequality? $$2(x-3)^2 < 32$$

Currently, I solve such inequalities by sketching. This just seems weird and I think there is much more to it than I know. What insights have you gained from thinking about such inequalities?

Answer 1

We notice that $2$ is a factor common to both the LHS and RHS. So we essentially reduced the problem to finding the solutions to the inequality: $$(x-3)^2 < 16$$ Noting that $16 = 4^2$, we bring it to the LHS. We then get, $$(x-3)^2 - 4^2 < 0$$ Applying the algebraic identity $a^2-b^2 = (a-b)(a+b)$ $$\implies (x-3-4)(x-3+4) < 0$$ $$\implies (x-7)(x+1) < 0$$ Since the product of these $2$ factors is negative, we must conclude that one of them must be negative while the other is positive. So, two cases are possible:

Case 1: $x-7 > 0 \implies x>7$ and $x+1<0 \implies x < -1$ (Not possible)

Case 2: $x-7 < 0 \implies x<7$ and $x+1>0 \implies x > -1$

Hence, via case 2, we conclude that the solution to the inequality is $x \in \ (-1,7) $

Answer 2

Try this $$ 2(x-3)^2 < 32 \\ (x-3)^2 < 16 \\ |x-3| < 4 \\ -4 < x-3 < 4 \\ -1 < x < 7. $$

The important lesson is: When you have $(x-3)^2 < 16$ and you "take square root of both sides", you do not get $x-3 < 4$.