## anonymous one year ago Algebra Help?

1. anonymous

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

2. anonymous

Rationalize the denominator and simplify.

3. anonymous

Why would it be the conjugate of the numerator and not the denominator? Shouldn't it be multiplied by $\frac{ \sqrt{a}+2\sqrt{y} }{ \sqrt{a}+2\sqrt{y} }$

4. rishavraj

u r correct

5. rishavraj

it should be $\frac{ \sqrt{a} + 2\sqrt{y} }{ \sqrt{a} - 2\sqrt{y}} \times \frac{ \sqrt{a} + 2\sqrt{y} }{ \sqrt{a} + 2\sqrt{y}}$

6. anonymous

Ok so that's the part I always have trouble with. I'm not the best at multiplying radicals.

7. rishavraj

$(a - b) \times (a + b) = a^2 - b^2$

8. rishavraj

so it would be $\frac{ (\sqrt{a} + 2\sqrt{y})^2 }{ (\sqrt{a})^2 - (2\sqrt{y})^2 }$

9. mathstudent55

|dw:1437103101604:dw|

10. anonymous

Ok so @rishavraj then its simplified to get $\frac{ a+2y }{ a-2y }$ or would those 2s be a 4s?

11. mathstudent55

|dw:1437103159002:dw|

12. rishavraj

$(2\sqrt{y})^2 = 4y ~~~~ and ~~~(a + b)^2 = a^2 + b^2 + 2ab$

13. rishavraj

and $(\sqrt{a} + 2\sqrt{y})^2 = (\sqrt{a})^2 + (2\sqrt{y})^2 + 2 \times \sqrt{a} \times 2\sqrt{y}$

14. anonymous

Why the $+2\times \sqrt{a}\times \sqrt{y}2$

15. anonymous

I mean $2\sqrt{y}$

16. rishavraj

(a + b)^2 = a^2 + b^2 + 2ab

17. anonymous

OK thanks I got it