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

<br /> a^2 = b^2<br />

From this statement alone it is not possible to claim that

<br /> a=b<br />

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

<br /> -4 \ne 4 = \sqrt{4 - 3(-4)}<br />

but

<br /> 16 = (-4)^2 = \left(\sqrt{4-3(-4)}\right)^2<br />

 

[Sheneron]

thank ye