Find the solution set for the inequality $\sqrt{a^{2} - 2a} > a - 4$

Question

All real values of $a$ for which $\displaystyle \sqrt{a^2-2a}>a-4$ , is

Here $\displaystyle \sqrt{a^2-2a}$ is defined

when $$\displaystyle a^2-2a\geq 0\Longrightarrow a\in(-\infty,0]\cup[2,\infty)\tag{1}$$

Also $\displaystyle a^2-2a>(a-4)^2$

$$\displaystyle a^2-2a>a^2+16-8a\Longrightarrow a\in(\tfrac{8}{3},\infty)\tag{2}$$

So we get answer is intersection of $(1)$ and $(2)$

So we get $\displaystyle a\in(\tfrac{8}{3},\infty)$

But answer given as $(-\infty,0]\cup [2,\infty)$

Please explain me what is wrong in my process, Thanks

Answer 1

That is because the RHS is non-positive when $a\in(-\infty,0]\cup[2,4]$.

Hence the solution set is given by: \begin{align*} S = (-\infty,0]\cup[2,4]\cup\left(\frac{8}{3},+\infty\right) = (-\infty,0]\cup[2,+\infty). \end{align*}

Hopefully this helps!

Answer 2

The domain gives: $a\geq2$ or $a\leq0$, which is valid for $a-4\leq0$ and we obtain: $2\leq a\leq4$ or $a\leq0$. But for $a>4$ we obtain: $$a^2-2a>(a-4)^2$$ or $$a>\frac{8}{3},$$ which gives $a>4$ only and we obtain the following answer: $$(-\infty,0]\cup[2,+\infty).$$

Answer 3

Following your process we have

• $\displaystyle \sqrt{a^2-2a}$ is defined when

$$\displaystyle a^2-2a\geq 0\Longrightarrow a\in D=(-\infty,0]\cup[2,\infty)\tag{1}$$

and this is fine. Now for $\sqrt{a^{2} - 2a} > a - 4$ we need to consider two cases (this is the part you have lost)

Case 1: $a-4<0 \iff a<4$

$$ \implies \sqrt{a^{2} - 2a} > a - 4 \;\;\text{always holds}$$

that is

$$a\in I_1=(-\infty,4)\tag{2}$$

Case 2: $a-4\ge 0 \iff a\ge 4$

$$ \sqrt{a^{2} - 2a} > a - 4\iff a^{2} - 2a>(a-4)^2$$

$$\displaystyle a^2-2a>a^2+16-8a\Longrightarrow a\in\left(\tfrac{8}{3},\infty\right)$$

that is

$$a\in I_2=[4,\infty)\tag{3}$$

So we get answer

$$D\cap(I_1\cup I_2)=D=(-\infty,0]\cup[2,\infty)$$

Answer 4

This answer provides a short summary of the general form of your inequality .

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Let $A(x)$ and $B(x)$ be some algebraic expressions . Then, we want to find the solution set of

$$\sqrt {A(x)}>B(x)$$

As you know, the above inequality is equivalent to :

$$ \begin{align}\begin{cases}A(x)≥0\\ B(x)<0\end{cases} \thinspace\thinspace\thinspace &{\color{red}{\textbf{and}}}\thinspace\thinspace\thinspace\begin{cases}B(x)≥0 \\ A(x)>B^2(x)\end{cases}\end{align} $$

which leads to the explicit solution set .

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$\rm{Comments :}$

By definition of the principal square root , we have $\sqrt {A(x)}≥0$ and $A(x)≥0$ and this justifies the first part of the solution set : $A(x)≥0\wedge B(x)<0$ .

If $B(x)≥0$, then we have consistent justification for squaring both sides of the inequality . Note that, in this case we don't need the restriction $A(x)>0$ . Because, $A(x)>B^2(x)≥0$ already implies that, $A(x)>0$ .