Posted by **Blair** on Friday, February 15, 2013 at 12:11pm.

Solve each quadratic in form equation.

1. 4y^4+9=13y^2

2. x-3x^1/22=0

3. (x-5)^2+2(x-5)-35=0

4. (x-2)^2-3x-2)+2=0

5. 2(x^2-5)^2-13(x^2-5)+20=0

6. 2x^2/3+5x^1/3-6=0

7. 6x^2/3-5x^1/3-6=0

8. x^-2+4x^-1=12

9. x^4-8x^2+7=0

10. x^4-2x^2-35=0

- Math -
**Steve**, Friday, February 15, 2013 at 12:42pm
just make a substitution using, say, u to get a quadratic.

For example, in #6, let u = x^1/3 and you have

2u^2+5u-6=0

u = 1/4 (-5 ± √73)

Since u=x^1/3, x = u^3, so

x = 1/16 (-305 ± 37√73)

Kinda nasty, but straightforward.

How about #5? Let u=x^2-5, to get

2u^2 - 13u + 20 = 0

(u-4)(2u-5) = 0

so, u=4,5/2

That means

(x^2-5) = 4 or (x^2-5) = 5/2

x^2=9 or x^2 = 15/2

x = ±3 or ±√(15/2)

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