Question

which expression should you multiply the numerator and denominator of 3 sqrt 3 / 3 sqrt 2x by to rationalize the denominator

Answer 1

Answer:

^3sqrt4x^2 and then ^3sqrt12x^2/2x

Step-by-step explanation:

Took the quiz and got it right

Answer 2

Answer:

$\sqrt{2x}$

Step-by-step explanation:

Given

$\frac{3\sqrt{3}}{3\sqrt{2x}}$

Required

Expression to rationalize the numerator and denominator of the above

To rationalize means to reduce a surd to a rational form by multiplying the surd by a suitable surd is know

To rationalize the above surd, we multiply the numerator and denominator by the surd expression at the denominator, which is $\sqrt{2x}$

Solving further to rationalize the given expression, we have

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

Multiply

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

Expand the expression to give

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

$\frac{3\sqrt{6x} }{3\sqrt{4x^2}}$

Simplify by dividing numerator and denominator by 3

$\frac{\sqrt{6x} }{\sqrt{4x^2}}$

Expand denominator

$\frac{\sqrt{6x} }{\sqrt{4} * \sqrt{x^2}}$

Square root of 4 is 2 and square root of $x^{2}$ is x, so we have

$\frac{\sqrt{6x} }{2 * x}$

$\frac{\sqrt{6x} }{2x}$

Hence, rationalizing $\frac{3\sqrt{3}}{3\sqrt{2x}}$ will give $\frac{\sqrt{6x} }{2x}$