Solve for $x$ in $x^2-5x+2\sqrt{x^2-5x+3}= 12$

Question

Solve for $x$ in $x^2-5x+2\sqrt{x^2-5x+3}= 12$

I've tried moving root term to one side and squaring both sides to get a $4$th-degree polynomial and find the roots that way. Is there any easier way of doing this?

Answer 1

One approach is to recognize the square of a sum and write $$x^2-5x+2\sqrt{x^2-5x+3}= 12\\ x^2-5x+3+2\sqrt{x^2-5x+3}+1= 16\\ (\sqrt{x^2-5x+3}+1)^2=16\\ \sqrt{x^2-5x+3}=3$$ Now square and you have a quadratic. You can justify taking only the positive root in the third to fourth line by looking at the left hand side.

Answer 2

A substitution can help here, but let's first move everything to one side:

$$x^2 - 5x - 12 + 2 \sqrt{x^2 - 5x + 3} = 0$$

Let $u= x^2 - 5x - 12$. Then $u+15 = x^2 - 5x + 3$ - this similarity in the radicand and the rest of the expression motivates this substitution, and our equation becomes

$$u + 2\sqrt{u+15} =0 $$

This is much easier to handle, as you might guess. We bring each term to separate sides and square them to get

$$u^2 = 4(u+15) = 4u+60 \implies u^2 - 4u - 60 = 0$$

Through some factoring,

$$(u-10)(u+6) = 0 \implies u = -6, u = 10$$

However, $u=10$ is an extraneous root from when we squared, so we ignore that. (Test it to see that it's extraneous!) From there, recalling our definition for $u$, this means that

$$x^2 - 5x - 12 = -6 \implies x^2 - 5x - 6 = 0 \implies (x-6)(x+1)=0 \implies x=-1, x=6$$

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Of course, Ross Millikan's answer is also valid (and by the looks of it, closer to the intentions of the exercise), but oh well.

Answer 3

There's a nice way of solving it. Let $y=\sqrt{x^{2}-5x+3}$. Then the equation becomes: $$y^{2}+2y-15=0$$ Which has roots $-5$ and $3$. If you are solving in real numbers, so the only possible answer is 3.

After that, just solve the equation $3^{2}=x^{2}-5x+3$ and you will get $x = -1$ and $x = 6$.

Remember that this is not a polynomial equation, it became a 4th degree polynomial equation only when you squared both sides. That does not mean that the roots of the new equation are roots of the original one. Even if you try putting $y=-5$, it will not satisfy the original equation. So, even if you consider complex numbers, there are only two roots.

When you square an equation, preferably you should check if the found roots are roots of the original equations.