# Solving an equation

1. Oct 2, 2008

### Sheneron

1. The problem statement, all variables and given/known data
$$x = \sqrt{4 - 3x}$$

3. The attempt at a solution
$$x^2 = 4 - 3x$$
$$x^2 + 3x - 4 = 0$$
$$(x+4)(x-1) = 0$$

$$x + 4 = 0$$
$$x = -4$$

$$x -1 = 0$$
$$x = 1$$

Plugging back into the original equation, -4 doesn't work and 1 does work. But, if I plug it into the quadratic they both work. So which is it?

2. Oct 2, 2008

### tiny-tim

Hi Sheneron!

(have a square-root: √ and a square: ² )

-4 would work if -4 = √(16).

But of course √ is defined to be ≥ 0.

So you "lose" any negative solutions.

(compare, for example, x = √1 and x² = 1 … they look the same, but the only solution to the first is x = 1, while the solution to the second is x = ±1 )

3. Oct 2, 2008

You must check the "solutions" you obtain in the original equation, not in one that comes from squaring that original equation. The reason is this: your two numbers come from a statement is

$$a^2 = b^2$$

From this statement alone it is not possible to claim that

$$a=b$$

automatically follows. Your result of $$x = -4$$ is an illustration of this: clearly

$$-4 \ne 4 = \sqrt{4 - 3(-4)}$$

but

$$16 = (-4)^2 = \left(\sqrt{4-3(-4)}\right)^2$$

4. Oct 2, 2008

thank ye