# Answer to Question #22492 in Trigonometry for Kristen Woods

Question #22492

Solve the given equations.

11a. x^(2/3)+6x^(1/3)=7

11b. (y+3)^2-8(y+3)+12=0

11a. x^(2/3)+6x^(1/3)=7

11b. (y+3)^2-8(y+3)+12=0

Expert's answer

11a. x^(2/3)+6x^(1/3)=7let t=x^(1/3)

t^2+6t-7=0

From Vieta's theorem

t1=1, t2=-7

x1=1^3=1, x2=(-7)^3=-343

x1=1 x2=-343

11b.(y+3)^2-8(y+3)+12=0

let y+3=t

t^2-8t+12=0

t^2-8t+16=4

(t-4)^2=4

t-4=+-2

t1=6, t2=2

then y1=3, y2=-1

t^2+6t-7=0

From Vieta's theorem

t1=1, t2=-7

x1=1^3=1, x2=(-7)^3=-343

x1=1 x2=-343

11b.(y+3)^2-8(y+3)+12=0

let y+3=t

t^2-8t+12=0

t^2-8t+16=4

(t-4)^2=4

t-4=+-2

t1=6, t2=2

then y1=3, y2=-1

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