Solve x = \sqrt{4 - 3x} Equation

[Sheneron]

Homework Statement

$$x = \sqrt{4 - 3x}$$

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?

 

[tiny-tim]

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?

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 )

 

[statdad]

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

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

 

[Sheneron]

thank ye