Solving for x: sqrt(x^2-x-10) = 10 + sqrt(x^2-11x)

[martinrandau]

I have a problem of where to start on this equation.

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

Solve for x.

 

[TD]

Square both sides, you'll get another square root on the RHS, isolate it on one side and square both sides again. Then you lost all square roots and you'll be able to solve.

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.

 

[bomba923]

Just expanding on what TD said,

\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

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\}

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

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