# Solving for x

• Sep 28th 2006, 03:02 PM
sarahh
Solving for x
I need to solve for x for these equations. Also, are any of these if and only if equations based on their solutions?
Code:

```1)  (3 - 2x)(x + 1) = x + 7 - 2(x^2 - 2) 2)  1 / (12 - 4x) = 1 / x^2 - 5x + 6 3)  4 - x = sqrt(x - 2) 4)  x(x + 7) = (x + 2)^2 + 3x - 4```
• Sep 28th 2006, 03:09 PM
Quick
Darn I forgot that 2 was negative :(
• Sep 28th 2006, 03:11 PM
topsquark
(3 - 2x)(x + 1) = x + 7 - 2(x^2 - 2)

3x + 3 - 2x^2 - 2x = x + 7 - 2x^2 + 4

3x + 3 - 2x = x + 11

x + 3 = x + 11

3 = 11

Thus, no such x exists.

-Dan

(Sorry Quick, on your line two: -2*-2 = + 4)
• Sep 28th 2006, 03:19 PM
topsquark
1 / (12 - 4x) = 1 / x^2 - 5x + 6

Can I presume that the RHS is 1/(x^2 - 5x + 6) ?

Multiply both sides by 12 - 4x.

1 = (12 - 4x)/(x^2 - 5x + 6)

Multiply both sides by x^2 - 5x + 6.

x^2 - 5x + 6 = 12 - 4x

We can make this easier on ourselves by doing a bit of factoring here. Otherwise we can bring everything over to the LHS and solve from there. However:

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

Dividing both sides by x - 3 we get

x - 2 = -4

x = -2.

We need to check to be sure that x = -2 is an acceptable solution, that is that x = -2 doesn't produce a 0 in either of the denominators in the original problem. It doesn't.

One more point, which is why I chose to do this this way. We need to check that x - 3 = 0 is not a solution. (This is the factor we cancelled. You always need to be careful of this.) x = 3 is NOT a solution because it produces a 0 in one of the denominators of the original expression (in fact both of them.)

So x = -2 is the only solution.

-Dan
• Sep 28th 2006, 03:23 PM
topsquark
4 - x = sqrt(x - 2)

Square both sides:

(4 - x)^2 = x - 2

16 - 8x + x^2 = x - 2

x^2 - 9x + 18 = 0

(x - 6)(x - 3) = 0

So x - 6 = 0 => x = 6
Or x - 3 = 0 => x = 3

Now, whenever you square both sides of an equation during a solution you should always check to make sure any solutions you get are correct. Squaring is an operation that occasionally gives extra solutions.

I note that x = 6 gives -2 = 2 in the original equation so this is not a solution.

x = 3 checks, so this is the solution.

-Dan
• Sep 28th 2006, 03:25 PM
topsquark
x(x + 7) = (x + 2)^2 + 3x - 4

x^2 + 7x = x^2 + 4x + 4 + 3x - 4

x^2 + 7x = x^2 + 7x

0 = 0

So we see that any value of x satisfies this equation.

-Dan
• Sep 28th 2006, 03:31 PM
topsquark
Quote:

Originally Posted by sarahh
Also, are any of these if and only if equations based on their solutions?

Any equation will define the "if" part. For example: "4 - x = sqrt(x - 2)" implies that "x = 3."

As far as the "only if" part is concerned that would be a statement like: "x = 3" implies "4 - x = sqrt(x - 2)." This is also true.

Basically any equation with a solution for x will have the "if and only if" property.

We can make a similar argument for equation 4 so say that it does as well.

Does equation 1? I mean "(3 - 2x)(x + 1) = x + 7 - 2(x^2 - 2)" implies that "no x exists" and "no x exists" implies "(3 - 2x)(x + 1) = x + 7 - 2(x^2 - 2)." This could be considered to be an "if and only if" but at the same time both statements are empty of meaning because there is no such x. Some would say that this is a biconditional statement that is true "vacuously." That is, the statement can be considered to be true because there is no example that contradicts it. (Though there is also no example that satisfies it, either.)

-Dan