# Answer to Question #20255 in Trigonometry for Kristen Woods

Question #20255

solve each quadratic in form equation

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

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

Expert's answer

let x^2-5=t then

2t^2-13t+20=0

D=13^2-4*2*20=9

t=(13+-3)/4

t=4 or t=5/2x^2=t+5

so

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

x1=-3, x2=3, x3=-sqrt(15/2), x4=sqrt(15/2)

2t^2-13t+20=0

D=13^2-4*2*20=9

t=(13+-3)/4

t=4 or t=5/2x^2=t+5

so

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

x1=-3, x2=3, x3=-sqrt(15/2), x4=sqrt(15/2)

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