Question

what is the simplest form of square root of

Answer 1

First step is to simplify the inside of the radical. There are x's on both top and bottom of the fraction. So we subtract their exponents. x³ - x¹ = x². Because the x³ was in the denominator and it was bigger, the x² goes on the bottom.

Your new equation is:

$\sqrt{\frac{126y^{5}}{32x^{2}} }$

Now, I'd suggest factoring each term to correspond with the square root.

126y⁵ = 9 · 14 · y² · y² · y

32x² = 4 · 4 · 2 · x²

9, y², and y² can be square rooted in the numerator. 4, 4, and x² in the denominator. So we pull out their square roots and place them outside of the radical. The rest stays in the radical:

$\frac{3 * y * y}{2*2*x}\sqrt{\frac{14y}{2}}$

Simplify to get your final answer:

$\frac{3y^{2}}{4x} \sqrt{7y}$

It's easier to explain on paper and in person than over computer text, sadly. Hopefully you can follow along with this.

Answer 2

$\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\ -------------------------------$

$\bf \sqrt{\cfrac{126xy^5}{32x^3}}\implies \sqrt{\cfrac{126}{32}\cdot \cfrac{xy^5}{x^3}}\implies \sqrt{\cfrac{63}{16}\cdot \cfrac{y^5}{x^3x^{-1}}}\implies \sqrt{\cfrac{63y^{4+1}}{16x^{3-1}}} \\\\\\ \sqrt{\cfrac{(9\cdot 7)y^4y^1}{(4\cdot 4)x^2}}\implies \sqrt{\cfrac{3^2\cdot 7(y^2)^2y}{4^2x^2}}\implies \cfrac{\sqrt{3^2\cdot 7(y^2)^2y}}{\sqrt{4^2x^2}} \\\\\\ \cfrac{3y^2\sqrt{7y}}{4x}$