SOLUTION: Solve the each of the following equations algebraically. A. {{{ 2(x - 5)(2x + 3) + 2(x - 5)^2 = 0 }}} B. {{{ 2x^(2/3) - 5x^(1/3) = 12 }}} C.

Algebra ->  Finance -> SOLUTION: Solve the each of the following equations algebraically. A. {{{ 2(x - 5)(2x + 3) + 2(x - 5)^2 = 0 }}} B. {{{ 2x^(2/3) - 5x^(1/3) = 12 }}} C.      Log On


   

Question 906749: Solve the each of the following equations algebraically.
A. +++2%28x+-+5%29%282x+%2B+3%29+%2B+2%28x+-+5%29%5E2+++=+++0+++
B. +++2x%5E%282%2F3%29++++-++++5x%5E%281%2F3%29++++=++++12++++

C. +++%28y++-++2%29%5E%281%2F2%29+++-+++%285y+%2B+1%29%5E%281%2F2%29++=++-++3++

Please explain how to solve
Thank you
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
A. +++2%28x+-+5%29%282x+%2B+3%29+%2B+2%28x+-+5%29%5E2+++=+++0+++
+++2%28x+-+5%29%28%282x+%2B+3%29+%2B+%28x+-+5%29%29+++=+++0+++
2%28x-5%29%282x%2B3%2Bx-5%29=0
%28x-5%29%283x-2%29=0
x=5 OR x=2/3


B. +++2x%5E%282%2F3%29++++-++++5x%5E%281%2F3%29++++=++++12++++
x^3*2x^(2/3)/x^3 +5x^3*x^(1/3)/x^3= 12
2x^6+5x^9=12x^3
5x^9+2x^6-12x^3=0
x^3(5x^6+2x^3-12)=0
x=0 is one solution
lt x^3=a
5a^2+2a-12=0
Find the roots of the equation by quadratic formula

a= 5 b= 2 c= -12

b^2-4ac= 4 - -240
b^2-4ac= 244 sqrt%28%09244++++%09%29= 15 5/8
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -2 + 15 5/8 )/ 10
x1= 1 3/8
x2=( -2 -15 5/8 ) / 10
x2= -1 3/4
a=11/8 OR -7/4
x^3=11/8 OR x^3= -7/4





C. +++%28y++-++2%29%5E%281%2F2%29+++-+++%285y+%2B+1%29%5E%281%2F2%29++=++-++3++