SOLUTION: solve equation, by making appropriate substitution. a)x^-2+x^-1-56=0 U= U^2= b)x^2/3-x^1/3-12=0 U= U^2=

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: solve equation, by making appropriate substitution. a)x^-2+x^-1-56=0 U= U^2= b)x^2/3-x^1/3-12=0 U= U^2=       Log On


   



Question 87941: solve equation, by making appropriate substitution.
a)x^-2+x^-1-56=0
U=
U^2=
b)x^2/3-x^1/3-12=0
U=
U^2=

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
solve equation, by making appropriate substitution. 

Rule: let U equal the variable part of the second term (without the coefficient)
Then UČ will equal the variable part of the first term (without the coefficient)

a)x%5E%28-2%29%2Bx%5E%28-1%29-56=0

U+=+x%5E%28-1%29, which you see is the variable part of the second term

U%5E2=+%28x%5E%28-1%29%29%5E2+=+x%5E%28-2%29, which is the variable part of the first term 

x%5E%28-2%29%2Bx%5E%28-1%29-56=0 becomes

U%5E2%2BU%5E2-56=0

%28U%2B8%29%28U-7%29=0
U%2B8=0,  U-7=0
U=-8,   U=7

Now substitute back, using U+=+x%5E%28-1%29,

x%5E%28-1%29+=+-8,  x%5E%28-1%29=7

Use the definition of a negative exponent to rewrite those:

1%2Fx+=+-8,  +1%2Fx+=+7

1%2Fx+=+%28-8%29%2F1, 1%2Fx+=+7%2F1

Cross multiply:

-8x+=+1, 7x+=+1

x+=+-1%2F8, x+=+1%2F7

===================================
    
b)x%5E%282%2F3%29-x%5E%281%2F3%29-12=0
U=x%5E%281%2F3%29,  the variable part of the second term
U%5E2=+%28x%5E%281%2F3%29%29%5E2+=+x%5E%282%2F3%29,  the variable part of the first term 

x%5E%282%2F3%29-x%5E%281%2F3%29-12=0 becomes

U%5E2-U-12=0

%28U-4%29%28U%2B3%29=0
U-4=0,  U%2B3=0
U=4,   U=-3

Now substitute back, using U+=+x%5E%281%2F3%29,

x%5E%281%2F3%29+=+4,  x%5E%281%2F3%29=-3

Use the definition of a fractional exponent to rewrite those:

root%283%2Cx%29+=+4,  root%283%2Cx%29+=+-3

Cubing both sides of both equations:

%28root%283%2Cx%29%29%5E3+=+4%5E3, %28root%283%2Cx%29%29%5E3+=+%28-3%29%5E3

x+=+64, x+=+-27

Edwin