I have the equation $2+\sqrt{12-2x}=x$ and my textbook tells me that the solution is $4$.
I arrived at a non real number solution so must have gone way of course somewhere:
$2+\sqrt{12-2x}=x$
$2^2+12-2x=x^2$ # square both sides to get rid of the radical
$x^2-2x+16=0$ # rearrange
Cannot see a way of solving this quadratic via factoring, so went with quadratic formula:
$a=1$, $b=-2$, $c=16$
$$\frac{2\pm\sqrt{-2^2-4(1)(16)}}{2(1)}$$
$$\frac{2\pm\sqrt{4-64}}{2}$$
$$\frac{2\pm\sqrt{-60}}{2}$$
$$\frac{2\pm\sqrt{4}\sqrt{15}(i)}{2}$$
$$\frac{2\pm2i\sqrt{15}}{2}$$
First solution:
$$\frac{2}{2}+\frac{2i\sqrt{15}}{2} = 1+\frac{i\sqrt{15}}{2}$$
Second solution:
$$\frac{2}{2}-\frac{2i\sqrt{15}}{2} = 1-i\sqrt{15}$$
Where did I go wrong and how can I arrive at the provided solution $4$?
Best Answer
Writing your equation in the form $$\sqrt{12-2x}=x-2$$ and squaring $$12-2x=x^2-4x+4$$ collecting like terms $$x^2-2x-8=0$$ Can you solve this equation?