How do you solve $\sqrt{x + 2} - x = 0$ $sqrt(x+2)-x=0$ ?

Question

How do you solve $\sqrt{x + 2} - x = 0$ $sqrt(x+2)-x=0$ ?

1 Answer

Impact of this question

12415 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-17T12:16:37

$x = 2$ $x=2$

Explanation:

Rearrange to get the square root on one side on its own:
$\sqrt{x + 2} - x = 0$ $sqrt(x+2)-x=0$
$\sqrt{x + 2} = x$ $sqrt(x+2)=x$

Square it out to get a quadratic:
$x + 2 = x^{2}$ $x+2=x^2$
$x^{2} - x - 2 = 0$ $x^2-x-2=0$

Quadratic formula:
$x = \frac{1}{2} (1 \pm \sqrt{1 + 8})$ $x=1/2(1+-sqrt(1+8))$
$x = \frac{1}{2} (1 \pm 3)$ $x=1/2(1+-3)$
$x = - 1, 2$ $x=-1,2$

But when we square in our working we need to check that we haven't accidentally introduced extra roots. In this case we have: -1 is not a solution of the original equation, although 2 is.

So there is one solution: $x = 2$ $x=2$ .

Science

Math

Humanities

... and beyond

How do you solve $\sqrt{x + 2} - x = 0$ $sqrt(x+2)-x=0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve √x+2−x=0sqrt(x+2)-x=0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $\sqrt{x + 2} - x = 0$ $sqrt(x+2)-x=0$ ?