Solve an inequality involving two square roots by comparing radicands

Key takeaway: Because the square-root function is increasing on its domain, you can compare what is inside the radicals once both expressions are defined. The domain check comes first; skipping it is where many students lose points.

We want to solve

$\sqrt{3x+1}>\sqrt{x-5}.$

Step 1: Check the domain

For the square roots to be real, we need

$3x+1\ge 0\quad\text{and}\quad x-5\ge 0.$

So

$x\ge -\frac13$

and

$x\ge 5.$

Together, the domain is

$x\ge 5.$

Step 2: Use monotonicity

Since $\sqrt{x}$ is increasing, on the domain we can compare the radicands:

$3x+1>x-5.$

Solve it:

$2x>-6$

so

$x>-3.$

Step 3: Combine with the domain

Intersect with $x\ge 5$:

$x\ge 5.$

Therefore the solution set is

$\boxed{[5,\infty)}.$

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Pitfalls the pros know 👇 A very common mistake is to square both sides immediately and forget to check the domain. Here that would be risky, because square roots only make sense for nonnegative radicands. Also, the strict inequality stays strict, so do not accidentally write $\ge$ after squaring.

What if the problem changes? If the inequality were reversed,

$\sqrt{3x+1}<\sqrt{x-5},$

the domain would still be $x\ge 5$, but the comparison of radicands would give $3x+1<x-5$, which leads to $x<-3$. Intersecting with the domain would then give no real solution.

Tags: domain of radicals, monotonic function, radical inequality