# Answer to Question #22212 in Algebra for Brenden Fields

Question #22212

Solve each quadratic in form equation.

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

6. 2x^(2/3)+5x^(1/3)=12

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

6. 2x^(2/3)+5x^(1/3)=12

Expert's answer

if (x^2 - 5) = y

2y^2-13y+20=0

y=1/(2*2)*(13+-Sqrt[13^2-4*2*20])= 1/4 * (13+-3)

y1 = 4

y2 = 2.5

& (x^2 - 5) = 4

x^2=9

x=+-3

(x^2 - 5) = 2.5

x^2=7.5

x=+-Sqrt[7.5]

Answer: x = { 3, -3, Sqrt[7.5], -Sqrt[7.5]

x^(1/3)=y

2y^2+5y-12=0

y=1/(2*2)*(-5+-Sqrt[5^2+4*2*12]=1/4*(-5+-11)

y1=-4

y2=3/2

x^(1/3)=-4

x=-64

x^(1/3)=3/2

x=81/8

Answer: x = -64, 81/8

2y^2-13y+20=0

y=1/(2*2)*(13+-Sqrt[13^2-4*2*20])= 1/4 * (13+-3)

y1 = 4

y2 = 2.5

& (x^2 - 5) = 4

x^2=9

x=+-3

(x^2 - 5) = 2.5

x^2=7.5

x=+-Sqrt[7.5]

Answer: x = { 3, -3, Sqrt[7.5], -Sqrt[7.5]

x^(1/3)=y

2y^2+5y-12=0

y=1/(2*2)*(-5+-Sqrt[5^2+4*2*12]=1/4*(-5+-11)

y1=-4

y2=3/2

x^(1/3)=-4

x=-64

x^(1/3)=3/2

x=81/8

Answer: x = -64, 81/8

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