Asked by Kristen
I have to solve each equation using a u substitution and need help understanding how to do this.
The first equation is:
1. x^2/3-2x^1/3-15=0
2. x+4=7 Squareroot x+4
The first equation is:
1. x^2/3-2x^1/3-15=0
2. x+4=7 Squareroot x+4
Answers
Answered by
Reiny
x^2/3-2x^1/3-15=0
(x^(1/3))^2 - 2x^(1/3) - 15 = 0
suppose we let x^(1/3) = y, then we get
y^2 - 2y - 15 = 0
(y-5)(x+3) = 0
y = 5 or y = -3
then x^(1/3)= 5 , cube both sides
x = 125
or
x^(1/3) = -3
x = -27
2.
x+4 = 7√(x+4)
square both sides
x^2 + 8x + 16 = 49(x+4)
x^2 + 8x + 16 = 49x +196
x^2 - 41x - 180=0
(x-45)(x+4) = 0
x = 45 or x = -4
BUT, since we squared, we must verify all answers
if x = 45
LS = 45+4 = 49
RS = 7√49 = 49 , works
if x = -4
LS = 0
RS = 7√0 = 0 , works
so x = 45 or x = -4
(x^(1/3))^2 - 2x^(1/3) - 15 = 0
suppose we let x^(1/3) = y, then we get
y^2 - 2y - 15 = 0
(y-5)(x+3) = 0
y = 5 or y = -3
then x^(1/3)= 5 , cube both sides
x = 125
or
x^(1/3) = -3
x = -27
2.
x+4 = 7√(x+4)
square both sides
x^2 + 8x + 16 = 49(x+4)
x^2 + 8x + 16 = 49x +196
x^2 - 41x - 180=0
(x-45)(x+4) = 0
x = 45 or x = -4
BUT, since we squared, we must verify all answers
if x = 45
LS = 45+4 = 49
RS = 7√49 = 49 , works
if x = -4
LS = 0
RS = 7√0 = 0 , works
so x = 45 or x = -4
Answered by
Kristen
Thank you so much this makes the examples provided make much more scense the way you broke them down! Thanks!
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