Math
posted by Blair .
Solve each quadratic in form equation.
1. 4y^4+9=13y^2
2. x3x^1/22=0
3. (x5)^2+2(x5)35=0
4. (x2)^23x2)+2=0
5. 2(x^25)^213(x^25)+20=0
6. 2x^2/3+5x^1/36=0
7. 6x^2/35x^1/36=0
8. x^2+4x^1=12
9. x^48x^2+7=0
10. x^42x^235=0

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+5u6=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^25, to get
2u^2  13u + 20 = 0
(u4)(2u5) = 0
so, u=4,5/2
That means
(x^25) = 4 or (x^25) = 5/2
x^2=9 or x^2 = 15/2
x = ±3 or ±√(15/2)