0 like 0 dislike
√a +2√y

Do you know how to rationalize the denominator and simply?

2 Answers

0 like 0 dislike
[tex]\frac{a+4\sqrt{ay}+y }{a-4y }[/tex]

Step-by-step explanation:

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

We know that:

-  [tex](a - b)*(a + b) = a^{2} -b^{2}[/tex]
- we can always multiply by 1
- [tex]\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a} +2\sqrt{y} }=1[/tex]
- [tex](a+b)^2=a^2+2ab+b^2[/tex]


[tex]\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 }[/tex]
0 like 0 dislike
[tex]\frac{a+4\sqrt{ay}+4y }{a-4y}[/tex]


[tex]\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} }[/tex]

solve for:

Rationalized denominator

Step-by-step explanation:

1. Rationalize the denominator

[tex]\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} } * \frac{2\sqrt{y} }{2\sqrt{y} }[/tex]

2. Simplify

[tex]\frac{2\sqrt{y}(\sqrt{a}+2\sqrt{y} ) }{2\sqrt{y}(\sqrt{a}-2\sqrt{y}) }[/tex]

[tex]\frac{a+4\sqrt{ay}+4y }{a-4y}[/tex]
Welcome to AskTheTask.com, where understudies, educators and math devotees can ask and respond to any number related inquiry. Find support and replies to any numerical statement including variable based math, geometry, calculation, analytics, geometry, divisions, settling articulation, improving on articulations from there, the sky is the limit. Find solutions to numerical problems. Help is consistently 100 percent free!


No related questions found