Question

write the expression so the radicand has no perfect squares other than 1
\\(\sqrt{175x^9}\\)

\\(\sqrt{175x^9} = \square\\)
(type an exact answer, using radicals as needed.)

Explanation:

Step1: Factor the radicand

Factor \(175x^9\) into perfect square and non - perfect square parts. We know that \(175 = 25\times7\) and \(x^9=x^{8}\times x=(x^{4})^{2}\times x\). So \(\sqrt{175x^{9}}=\sqrt{25\times7\times x^{8}\times x}\)

Step2: Use the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (\(a\geq0,b\geq0\))

\(\sqrt{25\times7\times x^{8}\times x}=\sqrt{25}\times\sqrt{x^{8}}\times\sqrt{7x}\)

Step3: Simplify the perfect square roots

We know that \(\sqrt{25} = 5\) and \(\sqrt{x^{8}}=x^{4}\) (since \((x^{4})^{2}=x^{8}\)). So \(\sqrt{25}\times\sqrt{x^{8}}\times\sqrt{7x}=5\times x^{4}\times\sqrt{7x}=5x^{4}\sqrt{7x}\)

Answer:

\(5x^{4}\sqrt{7x}\)