SOLUTION: 1. Solve the equation: 2x^3 + 6x= 8x^2 2. Explain why it is incorrect to solve the equation (y-2) (y+2) =4 by setting y-2=2 and y+2=2 and solving the resulting equations. Tha

Algebra ->  Linear-equations -> SOLUTION: 1. Solve the equation: 2x^3 + 6x= 8x^2 2. Explain why it is incorrect to solve the equation (y-2) (y+2) =4 by setting y-2=2 and y+2=2 and solving the resulting equations. Tha      Log On


   

Question 145: 1. Solve the equation: 2x^3 + 6x= 8x^2
2. Explain why it is incorrect to solve the equation (y-2) (y+2) =4 by setting y-2=2 and y+2=2 and solving the resulting equations.
Thank you in advance.
Found 2 solutions by KMST, MathTherapy:
Answer by KMST(5484) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E3+%2B+6x=+8x%5E2-->2x%5E3+-+8x%5E2+%2B+6x+=+0-->2x%28x%5E2-4x%2B6%29=0
2x=0 --> highlight%28x=0%29 is a solution, but x%5E2-4x%2B6=0 has no real solutions,
because x%5E2-4x%2B6=%28x%5E2-4x%2B4%29%2B%286-4%29=%28x-2%29%5E2%2B2%3E=2 for every real x .
highlight%28x=0%29 is the only real solution.

%28y-2%29%28y%2B2%29=4 says that the product of two clearly different numbers is 4
The difference between y-2 and y%2B2 is y%2B2-%28y-2%29=y%2B2-y%2B2=4
y-2 and y%2B2 are different numbers and they cannot both be 2
2%2A2=4 but that says that a product of two identical numbers is 4
The two equations -->y=4 and y%2B2=2%7D%7D%3C--%3E%7B%7B%7By=2-2-->y=0 .
However, those solutions are not related to the original %28y-2%29%28y%2B2%29=4 equation, and solving them is like looking for the car keys under a street light at night, because you can see better there,
when you know that you accidentally dropped them on some buses in a shady spot.

We can find the solutions to the equation %28y-2%29%28y%2B2%29=4 .
%28y-2%29%28y%2B2%29=4<-->y%5E2-4=4-->y%5E2=4%2B4-->y%5E2=4%2A2-->

Answer by MathTherapy(10897) About Me  (Show Source):
You can put this solution on YOUR website!
1. Solve the equation: 2x^3 + 6x= 8x^2

2. Explain why it is incorrect to solve the equation (y-2) (y+2) =4 by setting y-2=2 and y+2=2 and solving the resulting equations.

Thank you in advance.
*********************
Tutor KMST(5456)'s claim,  that x = 0 is the only real solution, is FALSE!!

           2x%5E3+%2B+6x+=+8x%5E2
  2x%5E3+-+8x%5E2+%2B+6x+=+0
 2x%28x%5E2+-+4x+%2B+3%29+=+0 --- Factoring out GCF, 2x
2x(x - 3)(x - 1) = 0 --- Factoring TRINOMIAL
2x = 0    x - 3 = 0   x - 1 = 0 ----- Setting each factor = to 0