Solving $\sqrt{8+2x-x^2} > 6-3x$.

Question

So I was solving this inequality. $$\sqrt{8+2x-x^2} > 6-3x$$

• First things first, I obtained that the common domain of definition is actually $[-2,4]$.

• Next we would square and solve the quadratic that follows.

But the "solution" seems to have a part, where they took $6-3x \geq 0$,

which gave another restriction for $x$ as $(-\infty,2]$.

I did not understand this. Why was this necessary?

Answer 1

Let's say we have an inequality $\sqrt a>b$. We're often interested in getting rid of the square root, so we want to do something along the lines of 'squaring both sides'. But squaring both sides doesn't necessarily preserve the inequality.

Example:

$$\sqrt5>-3$$

But

$$\implies\sqrt5^2>(-3)^2 \implies 5\gt9$$

is clearly false.

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If you want a general rule, then you should use

$$|a|\gt|b|\iff a^2>b^2$$

This explains why you need to consider separate cases here

Answer 2

The domain gives $-2\leq x\leq4$.

For $x>2$ with the domain our inequality is true.

But for $x\leq2$ we can use squaring,

which gives $5x^2-19x+14<0$ or $1<x<2.8$ and since $(1,2]\cup(2,4]=(1,4]$, we get the answer: $$(1,4]$$

Answer 3

Just use these fundamental rules for irrational (in)equalities, valid on the domain $A\ge 0$: \begin{alignat}{2} &\bullet\quad \sqrt A> B\iff A>B^2\quad\text{OR}&&\quad B<0 \\ &\bullet\quad \sqrt A < B\iff A < B^2\quad\text{AND}&&\quad B\ge 0 \\ &\bullet\quad \sqrt A = B\iff A = B^2\quad\text{AND}&&\quad B\ge 0 \end{alignat}