$2+\sqrt{12-2x}=x$ solution is 4, cannot arrive at this solution

Question

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 $4$?

Answer 1

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?

Answer 2

When you square both sides you make a mistake $(2+\sqrt{12-2x})^2 \neq 2^2+ (\sqrt{12-2x})^2$!!!

How to do it: $$2+\sqrt{12-2x}=x$$

$$\sqrt{12-2x}=x-2$$

$$12-2x=(x-2)^2$$

$$12-2x=x^2-4x+4$$

$$x^2-2x-8=0$$

$$(x+2)(x-4)=0$$

So $x=4$ is a solution.

Answer 3

Hint

Avoid squaring which invites Extraneous Roots

Like Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$

let $y=\sqrt{12-2x}\ge0$

$x=\dfrac{12-y^2}2$

Solve the resulting quadratic equation in $x$

Answer 4

You fell for the "freshman's binomial": $(a+b)^2=a^2+b^2$.

Instead, you need to isolate the radical: $\sqrt{12-2x}=x-2$.

Then square both sides: $12-2x=(x-2)^2\implies 12-2x=x^2-4x+4\implies x^2-2x-8=0\implies (x-4)(x-2)=0\implies x=4,2$.

Then you need to check your solutions, since squaring both sides can introduce extraneous solutions:

$\sqrt{12-2\cdot 4}=4-2\iff \sqrt4=2\iff 2=2$. So $4$ is a solution.

And $\sqrt{12-2\cdot 2}=2-2\iff\sqrt8=0$. So $2$ is not a solution.

Answer 5

$$\sqrt{12-2x}=x-2,$$

$$12-2x=x^2-4x+4,$$

$$x^2-2x+1=9$$

and $x=1\pm3$.

$x=-2$ must be rejected.

Answer 6

The Eq. $$2+\sqrt{12-2x}=x$$, or $$\sqrt{12-2x}=x-2.$$ Let $x\ge2$ then only we can square both the sides and ger $$12-2x=(x-2)^2 \Rightarrow x^2-2x-8=0 \Rightarrow (x-4)(x+2)=00 \Rightarrow x=4 ~\mbox{or}~-2.$$ Since only $x=4$ satisfies the condition that $x \ge 2.$ So the only roots is $x=4$.

Answer 7

When you take $\sqrt{12-2x} = x -2$ and square both sides as has been suggested, which is a correct approach to the problem, you also incur the possibility that you will have an extraneous solution to the problem at hand by solving the new equation.

$$ 12 - 2x = x^{2} - 2x - 8$$

equtes to

$$ x^{2} - 2x -8 = (x-4)(x+2) = 0,$$

giving $x=-2;4$.

However, as $x=-2$ is not a solution of the original equation but $x=4$ is, it follows that

$$x=4$$

is the only solution to your problem.

Answer 8

Let $12-2x=4t^2$, and the equation reads

$$2+2t=6-2t^2$$ or $$t^2+t-2=0,$$ where $t$ is taken positive.

The solution is $t=1,x=4$.

Answer 9

$2+\sqrt{12-2x}$ decreases and $x$ increases.

Thus, our equation has one root maximum.

But $4$ is a root, which gives that it's an unique root of the equation.

Answer 10

Note that $(a + b)^{2} \neq a^{2} + b^{2}$.

So your approach to square both sides of the equation is not valid, because the left side $ 2 + \sqrt{12 - 2x} $, when squared, is not equal to $ 2^{2} + 12 - 2x $. Instead, use the FOIL (First, Outer, Inner, Last) method to square binomials such as $ 2 + \sqrt{12 - 2x} $.