Question

assuming x and y are both positive, write the following expression in simplest radical form. 2√(252x⁷y⁴)

Explanation:

Step1: Factor the radicand

First, factor 252, \(x^7\), and \(y^4\) into perfect squares and other factors.

• Factor 252: \(252 = 36\times7 = 6^2\times7\)
• Factor \(x^7\): \(x^7 = x^6\times x=(x^3)^2\times x\)
• Factor \(y^4\): \(y^4=(y^2)^2\)

So, \(\sqrt{252x^7y^4}=\sqrt{6^2\times7\times(x^3)^2\times x\times(y^2)^2}\)

Step2: Apply the square - root property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\))

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\), we can split the square root:
\(\sqrt{6^2\times7\times(x^3)^2\times x\times(y^2)^2}=\sqrt{6^2}\times\sqrt{(x^3)^2}\times\sqrt{(y^2)^2}\times\sqrt{7x}\)

Step3: Simplify the square roots of perfect squares

We know that \(\sqrt{a^2}=a\) for \(a\geq0\). So:

• \(\sqrt{6^2} = 6\)
• \(\sqrt{(x^3)^2}=x^3\) (since \(x>0\))
• \(\sqrt{(y^2)^2}=y^2\) (since \(y>0\))

Then \(\sqrt{6^2}\times\sqrt{(x^3)^2}\times\sqrt{(y^2)^2}\times\sqrt{7x}=6\times x^3\times y^2\times\sqrt{7x}=6x^3y^2\sqrt{7x}\)

Step4: Multiply by the coefficient outside the radical

The original expression is \(2\sqrt{252x^7y^4}\), so we multiply the result from step 3 by 2:
\(2\times6x^3y^2\sqrt{7x}=12x^3y^2\sqrt{7x}\)

Answer:

\(12x^{3}y^{2}\sqrt{7x}\)