Help with an equation?

ajlm said:
Okay, I've simplified the equation enough to get:

6=2x^2 - 3x

How do I get rid of that x^2 and it's coefficient, so I can solve for x?
Click to expand...

This is a quadratic equation - you do not get rid of the square term, per se.

write it as:

2x[sup:2y30f6dw]2[/sup:2y30f6dw] - 3x - 6 = 0

If you forgot how to solve quadratic equations, for a quick review, go to:

http://www.purplemath.com/modules/solvquad3.htm
 
ajlm said:
Okay, I've simplified the equation enough to get:
6=2x^2 - 3x
Click to expand...
That's not correct (but you're close). Should be 6 = 2x^2 - 11x

Now use the quadratic formula and you'll be ok.
 
ajlm said:
The original form of the equation is (x-2) / x = 4 - (x+4)/(x-3)

So, trying to solve it myself, I do come up with what you have: 2x^2 - 3x - 6, yet when I try to solve the quadratic (by completing the square and using the quadratic equation) my answer doesn't work in the original equation. so... either I'm solving for the roots incorrectly, or the quadratic (from the original) is wrong in the first place? Could someone point out what I'm doing wrong?
Click to expand...

If you multiply both sides of that equation by the least common denominator of all of the fractions, which is x(x - 3), you'll have this:

x(x - 3)*(x - 2)/x = x(x - 3)*4 - x(x - 3)*(x + 4)/(x - 3)

or,

(x - 3)(x - 2)= 4x*(x - 3) - x*(x + 4)

x[sup:3bkudbnm]2[/sup:3bkudbnm] - 5x + 6 = 4x[sup:3bkudbnm]2[/sup:3bkudbnm] - 12x - x[sup:3bkudbnm]2[/sup:3bkudbnm] - 4x

x[sup:3bkudbnm]2[/sup:3bkudbnm] - 5x + 6 = 3x[sup:3bkudbnm]2[/sup:3bkudbnm] - 16x

Let's get one side equal to 0 (I'm choosing to get 0 on the left side, so that the coefficient of the highest power of x will be positive.)

0 = 2x[sup:3bkudbnm]2[/sup:3bkudbnm] - 11x - 6

Now...can you take it from there? You can factor the non-zero side (Yes, it DOES factor), or you can use the quadratic formula.

Be sure to check your soultions to see if they make the original equation true.
 
Why are you trying to delete your posts? Your posts will help other students - and they will learn from these.
 
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