Question

√a +2√y √a-2√y Do you know how to rationalize the denominator and simply?

Answer 1

Answer:

$\frac{a+4\sqrt{ay}+4y }{a-4y}$

given:

$\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} }$

solve for:

Rationalized denominator

Step-by-step explanation:

1. Rationalize the denominator

$\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} } * \frac{2\sqrt{y} }{2\sqrt{y} }$

2. Simplify

$\frac{2\sqrt{y}(\sqrt{a}+2\sqrt{y} ) }{2\sqrt{y}(\sqrt{a}-2\sqrt{y}) }$

$\frac{a+4\sqrt{ay}+4y }{a-4y}$

Answer 2

Answer:

$\frac{a+4\sqrt{ay}+y }{a-4y }$

Step-by-step explanation:

As far as I understand, it looks like this:   $\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} }$

We know that:

1. $(a - b)*(a + b) = a^{2} -b^{2}$  
2. we can always multiply by 1
3. $\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a} +2\sqrt{y} }=1$  
4. $(a+b)^2=a^2+2ab+b^2$

Therefore,

$\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} } *1 =\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} } *\frac{\sqrt{a} +2\sqrt{y} }{\sqrt{a}+2\sqrt{y} } =\frac{(\sqrt{a}+2\sqrt{y})^2 }{(\sqrt{a})^2-(2\sqrt{y})^2 } =\frac{a+4\sqrt{ay}+y }{a-4y }$