- #1

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sqrt(x^2-x-10) = 10 + sqrt(x^2 - 11x)

Solve for x.

- Thread starter martinrandau
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- #1

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sqrt(x^2-x-10) = 10 + sqrt(x^2 - 11x)

Solve for x.

- #2

TD

Homework Helper

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Note: by squaring you may introduce new solutions. You'll have to check those, each expression under a root can't be negative, so cancel out false solutions.

- #3

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Just expanding on what TD said,

[tex] \sqrt {x^2 - x - 10} = 10 + \sqrt {x^2 - 11x} \Rightarrow x^2 - x - 10 = 100 + 20\sqrt {x^2 - 11x} + x^2 - 11x \Rightarrow [/tex]

[tex] x - 11 = 2\sqrt {x^2 - 11x} \Rightarrow 3x^2 - 22x - 121 = 0 \Rightarrow x = \frac{{22 \pm \sqrt {22^2 + 1452} }}{6} \Rightarrow x = \left\{ { - \frac{{11}}{3},11} \right\} [/tex]

However, you probably want *[tex] \boxed{x = 11} [/tex]* because [itex] -11 / 3 [/itex] won't work!

Why? Because squaring [itex] x - 11 = 2\sqrt {x^2 - 11x} [/itex] introduces false solutions!

[tex] \sqrt {x^2 - x - 10} = 10 + \sqrt {x^2 - 11x} \Rightarrow x^2 - x - 10 = 100 + 20\sqrt {x^2 - 11x} + x^2 - 11x \Rightarrow [/tex]

[tex] x - 11 = 2\sqrt {x^2 - 11x} \Rightarrow 3x^2 - 22x - 121 = 0 \Rightarrow x = \frac{{22 \pm \sqrt {22^2 + 1452} }}{6} \Rightarrow x = \left\{ { - \frac{{11}}{3},11} \right\} [/tex]

However, you probably want *[tex] \boxed{x = 11} [/tex]* because [itex] -11 / 3 [/itex] won't work!

Why? Because squaring [itex] x - 11 = 2\sqrt {x^2 - 11x} [/itex] introduces false solutions!

Last edited:

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