Question

Simplify 162x3y2−−−−−−−√ . Assume x and y are nonnegative.

Answer 1

Given expression: $\sqrt{162x^3y^2}$

In order to simplify given radical expression, let us convert it in three factors of radicals.

$\sqrt{162x^3y^2}=\sqrt{162}*\sqrt{x^3}*\sqrt{y^2}$

In order to simplify first part sqrt(162), we need to factor out greatest number that would be a perfect square.

162 = 81 * 2.

Therefore,

$\sqrt{162}=\sqrt{81 *2}$

We know, square root of 81 is 9.

Therefore, we can factor out 9 outside the radical.

$\sqrt{162}=\sqrt{81 *2}=9\sqrt{2}$

Let us simplify sqrt(x^3) now.

$\sqrt{^3}=\sqrt{x*x*x}=x\sqrt{x}$    Each pair of a factor of a variable or number inside square root get a factor out.

$\sqrt{y^2}=y$

Let us combine the solution:

$\sqrt{162x^3y^2}=9\sqrt{2} *x\sqrt{x}*y=9xy\sqrt{x}$

Therefore, final answer is $9xy\sqrt{x}$.

Answer 2

Answer:

$9xy\sqrt{2x}$

Step-by-step explanation:

Given expression is,

$\sqrt{162x^3y^2}$

$=(162x^3y^2)^\frac{1}{2}$    $(\because \sqrt[n]{x} =x^\frac{1}{n})$

$=162^\frac{1}{2}(x^3)^\frac{1}{2} (y^2)^\frac{1}{2}$  $(\because (ab)^m=a^mb^m)$

$=(81\times 2)^\frac{1}{2} x^{3\times \frac{1}{2}} y^{2\times \frac{1}{2}}$

$=(81)^\frac{1}{2} (2)^\frac{1}{2} x^{\frac{3}{2}} y^{\frac{2}{2}}$

$=9(2)^\frac{1}{2} x^{1+\frac{1}{2}} y^1$

$=9(2)^\frac{1}{2} x^{1}.x^{\frac{1}{2}} y$   $(\because a^{m+n}=a^m.a^n )$

$=9xy(2x)^\frac{1}{2}$

$=9xy\sqrt{2x}$