How do you solve $\sqrt{x + 6} = x$ $sqrt(x + 6) = x$ and find any extraneous solutions?

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How do you solve $\sqrt{x + 6} = x$ $sqrt(x + 6) = x$ and find any extraneous solutions?

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Answer 1 · 2017-10-09T13:19:54

$x = 3, x = - 2 is extraneous$ $x=3,x=-2" is extraneous"$

Explanation:

$squaring both sides$ $color(blue)"squaring both sides"$

${(\sqrt{x + 6})}^{2} = x^{2}$ $(sqrt(x+6))^2=x^2$

$\Rightarrow x + 6 = x^{2}$ $rArrx+6=x^2$

$rearrange in standard form a x^{2} + b x + c = 0$ $"rearrange in standard form "ax^2+bx+c=0$

$\Rightarrow x^{2} - x - 6 = 0$ $rArrx^2-x-6=0$

$the factors of - 6 which sum to - 1 are - 3 and + 2$ $"the factors of - 6 which sum to - 1 are - 3 and + 2"$

$\Rightarrow (x - 3) (x + 2) = 0$ $rArr(x-3)(x+2)=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$x - 3 = 0 \Rightarrow x = 3$ $x-3=0rArrx=3$

$x + 2 = 0 \Rightarrow x = - 2$ $x+2=0rArrx=-2$

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

$substitute each of the possible solutions into the$ $"substitute each of the possible solutions into the"$
$original equation to test their validity$ $"original equation to test their validity"$

$x = 3 \to \sqrt{3 + 6} = \sqrt{9} = 3 = x \Rightarrow valid solution$ $x=3tosqrt(3+6)=sqrt9=3=xrArr" valid solution"$

$x = - 2 \to \sqrt{- 2 + 6} = \sqrt{4} = 2 \neq x \Rightarrow extraneous$ $x=-2tosqrt(-2+6)=sqrt4=2!=xrArr"extraneous"$

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How do you solve $\sqrt{x + 6} = x$ $sqrt(x + 6) = x$ and find any extraneous solutions?

1 Answer

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Humanities

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How do you solve √x+6=xsqrt(x + 6) = x and find any extraneous solutions?

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Explanation:

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How do you solve $\sqrt{x + 6} = x$ $sqrt(x + 6) = x$ and find any extraneous solutions?