Answer to Question #22205 in Trigonometry for Brenden Fields

Question

Solve each quadratic in form equation.

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

2. x-3x^(1/2)+2=0

Answer 1

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

let t=y^2 then 4t^2+9=13t
4t^2-13t+9=0
D=13^2-4*4*9=169-144=25
t=(13+- sqrt(25))/8=(13+- 5)/8
so t=1 or t=9/4

then y^2=1 or y^2=9/4
so we get y1=-1, y2=1, y3=-3/2, y4=3/2

2. x-3x^(1/2)+2=0

let t=x^(1/2) then t^2-3t+2=0
from Vieta's theorem t1=1, t2=2
x^(1/2)=1 or x^(1/2)=2
Solution: x1=1 , x2=4