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The above question is from Serge Lang's basic mathematics. The question asks if there are any values of x which satisfy the above equation.

Serge Lang's answer key states that there is no solution.

From the equation, I squared both sides, then used the quadratic formula to find $\sqrt{ x} = 2-\sqrt{1/3}$ or $2+\sqrt{1/3}$. I substituted the former into the equation, and found that it would equate $\sqrt{x-2}$ to a negative answer, and hence is wrong since $\sqrt{x-2}$ refers to the positive root of $x-2$. For the second answer, I substituted and found that both sides of the equation equaled out. I drew $y=\sqrt{x-2}$ and $y=\sqrt{x}$ in a graphing calculator, and found they intersected correctly at a point.

Yet the book states there are no solutions. What went wrong?

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  • $\begingroup$ Either you copied the equation wrong or the book is wrong, for the equation you give definitely has a real number solution. $\endgroup$ – rogerl May 14 '16 at 2:21
  • $\begingroup$ A small trick when solving this type of problem, the answer found with the minus is always wrong, whereas the answer with the plus is usually right, when you use the quadratic formula. $\endgroup$ – Simply Beautiful Art May 14 '16 at 10:40
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Alpha finds a solution $x=\frac {13}3 - \frac 4{\sqrt 3}$ and shows a graph like you drew, so it appears the answer key is wrong.

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Set $\sqrt{x}=t$, so the equation becomes $\sqrt{t^2-2}=3-2t$. We have some conditions, though: we need $t\ge0$, $t^2\ge2$ and $3-2t\ge0$, that can be put together as $\sqrt{2}\le t\le 3/2$.

Now we can square and get $t^2-2=9-12t+4t^2$, so the quadratic $$ 3t^2-12t+11=0 $$ whose roots are $$ \frac{6-\sqrt{3}}{3},\qquad\frac{6+\sqrt{3}}{3} $$ Note that $$ \sqrt{2}<\frac{6-\sqrt{3}}{3}<\frac{3}{2}<\frac{6+\sqrt{3}}{3} $$ so $t=(6-\sqrt{3})/3$ is a solution, that translates to $$ x=\left(\frac{6-\sqrt{3}}{3}\right)^2=\frac{13-4\sqrt{3}}{3} $$

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$$\sqrt{x-2}+2\sqrt x=3$$ $x\ge2$ $$x-2+4\sqrt {x(x-2)}+4x=9$$ $$4\sqrt {x(x-2)}=11-5x$$ $x\le\frac{11}5$ $$16x(x-2)=121-110x+25x^2$$ $$9x^2-78x+121=0$$ $$x_{1,2}=\frac{13}{3}\pm\frac{4}{\sqrt3}$$ But $x_{1}=\frac{13}{3}+\frac{4}{\sqrt3}>\frac{11}5$ $$2<x_2=\frac{13}{3}+\frac{4}{\sqrt3}<\frac{11}5$$ Then $x_2=\frac{13}{3}-\frac{4}{\sqrt3} -$ solution of the $\sqrt{x-2}+2\sqrt x=3$

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  • $\begingroup$ It looks like you meant $x_2 = \frac{13}{3} \color{red}{-} \frac{4}{\sqrt{3}}$. $\endgroup$ – N. F. Taussig May 14 '16 at 9:41

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