On the number line, the point \(x\) is \(\sqrt{13}\) units to the right of \(-4\). In which interval does \(x\) lie?

Correct answer: Between \(-1\) and \(0\)

Moving \(\sqrt{13}\) units to the right from \(-4\) gives \(x=-4+\sqrt{13}\). Since \(3^2<13<4^2\), we have \(3<\sqrt{13}<4\). Adding \(-4\) throughout gives \(-1<x<0\), so \(x\) lies between \(-1\) and \(0\). Option B is incorrect because the value of \(x\) is still negative. Exam tip: locate a square root between two consecutive integers before adding or subtracting it.