How do I find the extraneous solution of $\sqrt{x + 4} = x - 2$ $sqrt(x+4)=x-2$ ?

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How do I find the extraneous solution of $\sqrt{x + 4} = x - 2$ $sqrt(x+4)=x-2$ ?

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Answer 1 · 2018-03-08T22:41:48

$x = 0 is an extraneous solution$ $x=0" is an extraneous solution"$

Explanation:

$square both sides to 'undo' the radical$ $"square both sides to 'undo' the radical"$

${(\sqrt{x + 4})}^{2} = {(x - 2)}^{2}$ $(sqrt(x+4))^2=(x-2)^2$

$\Rightarrow x + 4 = x^{2} - 4 x + 4$ $rArrx+4=x^2-4x+4$

$rearrange into standard form$ $"rearrange into standard form"$

$\Rightarrow x^{2} - 5 x = 0 \leftarrow in standard form$ $rArrx^2-5x=0larrcolor(blue)"in standard form"$

$\Rightarrow x (x - 5) = 0$ $rArrx(x-5)=0$

$\Rightarrow x = 0 or x = 5$ $rArrx=0" or "x=5$

$As a check$ $color(blue)"As a check"$

$Substitute these values into the original equation$ $"Substitute these values into the original equation"$

$x = 0 \to \sqrt{4} = 2 and 0 - 2 = - 2$ $x=0tosqrt4=2" and "0-2=-2$

$2 \neq - 2 \Rightarrow x = 0 is an extraneous solution$ $2!=-2rArrx=0color(red)" is an extraneous solution"$

$x = 5 \to \sqrt{9} = 3 and 5 - 2 = 3 \leftarrow True$ $x=5tosqrt9=3" and "5-2=3larr" True"$

$\Rightarrow x = 5 is the solution$ $rArrx=5color(red)" is the solution"$

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How do I find the extraneous solution of $\sqrt{x + 4} = x - 2$ $sqrt(x+4)=x-2$ ?

1 Answer

Explanation:

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How do I find the extraneous solution of √x+4=x−2sqrt(x+4)=x-2?

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How do I find the extraneous solution of $\sqrt{x + 4} = x - 2$ $sqrt(x+4)=x-2$ ?