How do you find extraneous solutions when solving radical equations?

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How do you find extraneous solutions when solving radical equations?

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Answer 1 · 2015-01-15T03:56:19

Let's start with an equation.

$x - 6 = \sqrt{x}$ $x-6=sqrtx$

This can be solved easily by:

$x^{2} - 12 x + 36 = x$ $x^2-12x+36=x$

$x^{2} - 13 x + 36 = 0$ $x^2-13x+36=0$

$(x - 9) (x - 4) = 0$ $(x-9)(x-4)=0$

$x = 9, x = 4$ $x=9, x=4$

but wait! we have to be careful whenever we square both sides of an equation.

Re-substitute the two answers into the original equation.

$9 - 6 = \sqrt{9}$ $9-6=sqrt9$
$3 = 3$ $3=3$

That's fine, one more.

$4 - 6 = \sqrt{4}$ $4-6 = sqrt4$
$- 2 \neq 2$ $-2 != 2$

Our extraneous answer is $x = 4$ $x=4$

To find the extraneous answer of any equation, just find the solutions and then re-substitute the solutions into the original equation. The answer that results in an illogical equation is the extraneous answer.

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How do you find extraneous solutions when solving radical equations?

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