Solution to $\sqrt{x-2} = 3- 2\sqrt{ x}$

Question

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?

Answer 1

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.

Answer 2

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 $$ x=\left(\frac{6-\sqrt{3}}{3}\right)^2=\frac{13-4\sqrt{3}}{3} $$

Answer 3

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