How do you solve $4 - x = \sqrt{x - 4}$ $4-x=sqrt(x-4)$ ?

Question

How do you solve $4 - x = \sqrt{x - 4}$ $4-x=sqrt(x-4)$ ?

1 Answer

Impact of this question

2032 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-17T09:08:11

$x = 4$ $x = 4$

Explanation:

The easy way

$\sqrt{x - 4}$ $sqrt(x-4)$ only takes Real values when $x \geq 4$ $x >= 4$ and then $\sqrt{x - 4} \geq 0$ $sqrt(x-4) >= 0$ .

When $x \geq 4$ $x >= 4$ then $4 - x \leq 0$ $4 - x <= 0$

So the only solution is $x = 4$ $x = 4$

The normal way

Square both sides of the equation to get:

$16 - 8 x + x^{2} = {(4 - x)}^{2} = {(\sqrt{x - 4})}^{2} = x - 4$ $16-8x+x^2 = (4-x)^2 = (sqrt(x-4))^2 = x - 4$

Note that squaring both sides of an equation may introduce spurious solutions (as it does in this case).

Subtract $x - 4$ $x - 4$ from both ends to get:

$x^{2} - 9 x + 20 = 0$ $x^2-9x+20 = 0$

This factors as:

$(x - 4) (x - 5) = 0$ $(x-4)(x-5) = 0$

So $x = 4$ $x = 4$ or $x = 5$ $x = 5$ .

Then check the solutions:

If $x = 4$ $x = 4$ then $4 - x = 0$ $4 - x = 0$ and $\sqrt{x - 4} = 0$ $sqrt(x-4) = 0$ . So $x = 4$ $x=4$ is a solution of the original equation.

If $x = 5$ $x = 5$ then $4 - x = - 1$ $4 - x = -1$ but $\sqrt{x - 4} = 1$ $sqrt(x - 4) = 1$ . So $x = 5$ $x=5$ is not a solution of the original equation.

Note that this happens because numbers have two square roots. When you square both sides of an equation you throw away some information.

Science

Math

Humanities

... and beyond

How do you solve $4 - x = \sqrt{x - 4}$ $4-x=sqrt(x-4)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 4-x=sqrt(x-4)4−x=√x−44-x=sqrt(x-4)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $4 - x = \sqrt{x - 4}$ $4-x=sqrt(x-4)$ ?