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October 1st, 2017, 12:13 PM   #1
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Question Simpler solve for this equation: sqrt(x^2-x-10)-sqrt(x^2-11x)=10

Hi, I have solved this equation but it seems like there might be a simpler way to do it (or it is not a simpler way at all ), but however the equation that I have solved looks like this:

$\displaystyle
\displaystyle \sqrt{x^2-x-10}-\sqrt{x^2-11x}=10\\\\\\\\
\displaystyle \sqrt{x^2-x-10}=10+\sqrt{x^2-11x}\\\\\\\\
\displaystyle x^2-x-10=100+20\sqrt{x^2-11x}+x^2-11x\\\\\\\\
\displaystyle 10x-20\sqrt{x^2-11x}=110\\\\\\\\
\displaystyle 10(x-2\sqrt{x^2-11x})=11\\\\\\\\
\displaystyle x-11=2\sqrt{x^2-11x}\\\\\\\\
\displaystyle x^2-22x+121=4(x^2-11)\\\\\\\\
\displaystyle x^2-22x+121=4x^2-44x\\\\\\\\
\displaystyle -3x^2+22x+121=0\\\\\\\\
$

Use the abc-formula
$\displaystyle \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\\\$
$\displaystyle
\displaystyle x=\frac{-22\pm\sqrt{(-22)^2-4\cdot -3\cdot 121}}{2\cdot -3}\\\\\\\\
\displaystyle x=\frac{-22\pm\sqrt{484+1452}}{-6}\\\\\\\\
\displaystyle x=\frac{-22\pm\sqrt{1936}}{-6}\\\\\\\\
\displaystyle x=\frac{-22\pm44}{-6}\\\\\\\\
$

The first solution (false)
$\displaystyle
\displaystyle x_1=\frac{-22+44}{-6}\\\\\\\\
\displaystyle x_1=\frac{22}{-6}\\\\\\\\
\displaystyle x_1=-\frac{11}{3} \ false\\\\\\\\
$

The second solution (true)
$\displaystyle
\displaystyle x_2=\frac{-22-44}{-6}\\\\\\\\
\displaystyle x_2=\frac{-66}{-6}\\\\\\\\
\displaystyle x_2=\frac{66}{6}\\\\\\\\
\displaystyle x_2=11 \ true\\\\\\\\
$

Last edited by DecoratorFawn82; October 1st, 2017 at 12:16 PM.
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October 1st, 2017, 02:35 PM   #2
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There is no simpler way. Squaring steps introduce spurious roots. Fourth line has an error, but it was corrected for the fifth line.
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October 1st, 2017, 07:03 PM   #3
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Multiplying both sides by $\sqrt{x^2 - x - 10} + \sqrt{x^2 - 11x} + 10$, gathering like terms and dividing by 10 leads to $x - 11 = 2\sqrt{x^2 - 11x}$, so $x^2 - 22x + 121 = 4x^2 - 44x$.
Hence $3x^2 - 22x + 121 = 0$, i.e. $(x - 11)(3x + 11) = 0$.
As $x = -11/3$ doesn't satisfy the original equation, $x = 11$ is the unique solution.
Thanks from DecoratorFawn82
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