# Inequality involving Square Root

$\sqrt{3x-x^2}<4-x$

I know I can't simply square both sides of an inequality. I have narrowed down the possible values of x => x belongs to [0,3] because the expression inside the square root cannot be negative. (Ignoring imaginary numbers)

And strangely enough, that is the given solution! => x belongs to [0,3]

Is the solution correct? And how? Doesn't the RHS matter?

• Like you correctly stated, the value under the square root must be positive ie $3x \ge x^{2} \implies x \in [0,3]$. Then, substitute in values of $x \in [0,3]$ to both sides and see if the inequality holds. – Mattos Apr 6 '15 at 7:09
• Now that you know that $x\in[0,3]$, you can square both sides of the inequality to create an equivalent inequality (since now you know both sides are non-negative). – user26486 Apr 6 '15 at 11:50
• Might be interesting to note that $x(3-x)$ is tangent to $4-x$ (at $(2,2)$). That means $x(3-x)\leq 4-x \forall x$... That they're tangent when $x(3-x)>1$ means the result will hold strictly when we apply the square root. – MichaelChirico Apr 6 '15 at 14:26

As you said, $x \in [0,3]$. Now if $x \in [0,3]$ the square root is well defined, and hence we have $0 \leq \sqrt {3x-x^{2}}< 4-x$. (You can square both sides since the square function is strictly increasing on $[0, \infty]$ Now squaring both sides we get $3x-x^{2}< x^{2}-8x+16$ hence $2x^{2}-11x+16>0$ but the discriminant of $2x^{2}-11x+16$ is strictly negative, and since the leading term $2>0$ then $2x^{2}-11x+16>0$ for all $x \in [0,3]$. Hence $[0,3]$ is the solution set.
The RHS matters, but in this case it just isn't any more restrictive on $x$ than the root is.