How do you solve $x^{2} + 2 = x + 4$ $x^2 + 2 = x + 4$ and find any extraneous solutions?

Question

How do you solve $x^{2} + 2 = x + 4$ $x^2 + 2 = x + 4$ and find any extraneous solutions?

1 Answer

Impact of this question

1277 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-20T10:17:02

$x = 2, - 1$ $x=2,-1$

Explanation:

$x^{2} + 2 = x + 4$ $x^2+2=x+4$

$x^{2} + 2 - x - 4 = 0$ $x^2+2-x-4=0$
Rearrange Equation

$x^{2} + 2 - x - 4 = 0$ $x^2+2-x-4=0$
Collect like terms so $2 - 4 = - 2$ $2-4=-2$

$x^{2} - 2 - x = 0$ $x^2-2-x=0$
Plug values into the Quadratic Formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \times 1 \times (- 2)}}{2 \times 1}$ $x=(-(-1)+-sqrt((-1)^2-4xx1xx(-2)))/(2xx1)$
Then solve
$x = \frac{1 \pm \sqrt{1 + 8}}{2}$ $x=(1+-sqrt(1+8))/(2)$

$x = \frac{1 \pm \sqrt{9}}{2}$ $x=(1+-sqrt(9))/(2)$

$x = \frac{1 \pm 3}{2}$ $x=(1+-3)/(2)$

$x = \frac{1 + 3}{2}$ $x=(1+3)/(2)$

$x = \frac{4}{2}$ $x=(4)/(2)$

$x = \frac{1 - 3}{2}$ $x=(1-3)/(2)$

$x = \frac{- 2}{2}$ $x=(-2)/(2)$

$x = 2, - 1$ $x=2,-1$

Science

Math

Humanities

... and beyond

How do you solve $x^{2} + 2 = x + 4$ $x^2 + 2 = x + 4$ and find any extraneous solutions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve x2+2=x+4x^2 + 2 = x + 4 and find any extraneous solutions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $x^{2} + 2 = x + 4$ $x^2 + 2 = x + 4$ and find any extraneous solutions?