SOLUTION: Consider the equation 10(x^2-49)/3x(x^2-4)(x+1)=0. Is x=7 permissible? Which values of x are excluded? Rewrite as a system of equations. Show your work.

Algebra ->  Expressions -> SOLUTION: Consider the equation 10(x^2-49)/3x(x^2-4)(x+1)=0. Is x=7 permissible? Which values of x are excluded? Rewrite as a system of equations. Show your work.      Log On


   



Question 1092687: Consider the equation 10(x^2-49)/3x(x^2-4)(x+1)=0. Is x=7 permissible? Which values of x are excluded? Rewrite as a system of equations. Show your work.
Found 2 solutions by addingup, greenestamps:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
10(x^2 - 49)/3x(x^2 - 4)(x + 1) = 0
Rewrite:
10/3x(x + 1)(x^2 - 49)(x^2 - 4) = 0
x(x + 1)(x^2 - 49)(x^2 - 4) = 0
x = 0; or x + 1 = 0; or x^2 - 49 = 0; or x^2 - 4 = 0
x = 0 or x = -1 or x^2 - 49 = 0 or x^2 - 4 = 0
x = 0 or x = -1 or (x - 7)(x + 7) = 0 or x^2 - 4 = 0
x = 0 or x = -1 or x - 7 = 0 or x + 7 = 0 or x^2 - 4 = 0
x = 0 or x = -1 or x = 7 or x = -7 or x^2 - 4 = 0
x = 0 or x = -1 or x = 7 or x = -7 or x^2 - 4 = 0
x = 0 or x = -1 or x = 7 or x = -7 or (x - 2)(x + 2) = 0
x = 0 or x = -1 or x = 7 or x = -7 or x - 2 = 0 or x + 2 = 0
x = 0 or x = -1 or x = 7 or x = -7 or x = 2 or x = -2

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!

When I copy the equation as you show it, this website interprets it as
10%28x%5E2-49%29%2F3x%28x%5E2-4%29%28x%2B1%29=0

However, the standard interpretation of the equation exactly as you show it would be
%2810%28x%5E2-49%29%2F3%29%28x%28x%5E2-4%29%28x%2B1%29%29=0

When you show "10(x^2-49)/3x..." the standard interpretation is that the "10(x^2-49) is divided by 3, and then the whole expression up to that point is multiplied by x and the other expressions that follow. To get everything following the "/" in the denominator of the fraction, you need to put everything following the "/"in parentheses: "10(x^2-49)/(3x...)".

So I am surprised the website interpreted your equation the way it did.

However, based on the way the question is worded, I suspect that is the intended equation.

The response you got previously used the standard interpretation, which makes a much less interesting problem than what I believe was intended.

So let's look at the equation as it was probably intended. We can factor the "x^2-49" in the numerator and the "x^2-4" in the denominator, so that all factors in both numerator and denominator are linear. Then we will be able to see everything we need to know about the equation.

10%28x%2B7%29%28x-7%29%2F3x%28x%2B2%29%28x-2%29%28x%2B1%29=0

The solutions are the values of x that make the numerator equal to 0; those values are -7 and 7.
The values of x that are excluded are the ones that make the denominator 0; those are 0, -2, 2, and -1.