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

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Answer 1 · 2017-05-20T10:17:02

$x=2,-1$

$x^2+2=x+4$

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

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

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

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-(-1)+-sqrt((-1)^2-4xx1xx(-2)))/(2xx1)$
Then solve
$x=(1+-sqrt(1+8))/(2)$

$x=(1+-sqrt(9))/(2)$

$x=(1+-3)/(2)$

$x=(1+3)/(2)$

$x=(4)/(2)$

$x=(1-3)/(2)$

$x=(-2)/(2)$

$x=2,-1$

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