# Answer to Question #22205 in Trigonometry for Brenden Fields

Question #22205

Solve each quadratic in form equation.

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

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

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

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

Expert's answer

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

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

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