## ksaimouli 3 years ago y^4+(1/16y^4)+1/2

1. ksaimouli

$y^4+\frac{ 1 }{ 16y^4 }+\frac{ 1 }{ 2 }$

2. ksaimouli

perfect square

3. ksaimouli

4. phi

you could make 16y^4 the common denominator and add up the terms the denominator is a perfect square. Does the numerator factor?

5. ksaimouli

is their any easy way instead of just taking common denominator

6. ksaimouli

@zepdrix

7. ksaimouli

i know a^2+2ab+b^2

8. phi

It is not that hard

9. phi

16 y^8 + 1 + 8 y^4

10. ksaimouli

=(a+b)^2

11. phi

re-arrange as 16 y^8 + 8 y^4 + 1 if you let y^4 = x, this is also 16x^2 + 8x +1

12. anonymous

i just tried$(y^2+\frac{1}{4y^2})^2$

13. phi

if it is square your only hope is (4x+1)^2 which works

14. ksaimouli

is that =0 because what happened to common denominator

15. ksaimouli

(4x^4+1)^2

16. anonymous

is it not true to say$(y^2+\frac{1}{4y^2})^2=y^4+2(y^2)(\frac{1}{4y^2})+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}$