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How do you solve and check for extraneous solutions in $x^2 + 2 = x + 4$ ?

1 Answer

Aug 7, 2015

Express the given equation in standard form, factor for solution values, and check each resulting value to ensure that it is not extraneous.
$x=2$ or $x=-1$

Explanation:

Given $x^2+2 = x+4$

Subtract $(x+4)$ from both sides to get into standard form
$color(white)("XXXX")$ $x^2-x-2=0$

Factor
$color(white)("XXXX")$ $(x-2)(x+1) = 0$

Either $(x-2)=0$ or $(x+1)=0$
$rArr$ $color(white)("XXXX")$ $x=2$ or $x=-1$

Checking $x=2$ by substituting into $x^2+2$ and $x+4$
$color(white)("XXXX")$ $x^2+2=(2)^2+2 = 6$
$color(white)("XXXX")$ $x+4=(2)+4 = 6$
$x^2+2=x+4$ so $x=2$ is not extraneous.

Similarly, checking $x=-1$ by substituting into $x^2+2$ and $x+4$
$color(white)("XXXX")$ $x^2+2=(-1)^2+2 = 3$
$color(white)("XXXX")$ $x+4=(-1)+4 = 3$
$x^2+2=x+4$ so $x=-1$ is not extraneous.

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