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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
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

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