# Solving x^2-6 = sqrt(x+6)

1. Feb 24, 2009

### rafehi

1. The problem statement, all variables and given/known data
x^2-6 = sqrt(x+6)
Solve for x.

3. The attempt at a solution

I've tried factorizing, but with no luck. I can get it down to:
x^4 - 12x^2 - x + 30 = 0,
but thinking back to my high school days, I'm sure there's an easier way to solve this question - just can't seem to remember how.

Any help?

2. Feb 24, 2009

### HallsofIvy

Fortunately, you already know that 3 is a solution, by "inspection". Dividing $x^4- 12x^2- x+ 30 by x- 3$ gives $x^3+ 3x^2- 3x- 10$ so $x^4- 12x^2- x+ 30= (x- 3)(x^3+ 3x^2- 3x- 10)$. If that cubic has any rational roots, they must evenly divide 10: so must be 1, -1, 2, -2, 5, -5, 10, or -10. Checking those, we find that x= -2 is a root! Dividing $x^3+ 3x^2- 3x- 10$ by x+ 2 gives $x^2+ x- 5$. Now you can use the quadratic formula to find the last two roots.

3. Feb 24, 2009

Er, $$-2$$ is not a solution to the original problem: the left hand side works out as $$-2$$ while the right hand side is $$2$$. I believe $$x = 3$$ is the only solution. HallofIvy didn't make that part clear, although I believe he meant it.

rafehi, did you find $$x = 3$$ or was it given as the answer? If it was given, and you need to find it, when you have

$$x^4- 12x^2- x+ 30 = 0$$

start by looking for rational zeros. Since the left side has two sign changes in coefficients, it has either two or no positive zeros. Try the integers first - you'll find 3. Divide by $$x - 3$$ and work with the cubic. Look at integer possibilities first again (simply because they are easiest to look for - just plug in the possibilities) - you'll find $$x = -2$$ is a solution to the derived equation. Divide the cubic by $$x + 2$$ to get the quadratic. Now use the quadratic formula to get the other values.

Since the four values come from an equation you obtained by squaring the initial problem each one must be checked in the original problem, and $$x = 3$$ is the only one that works there.

4. Feb 24, 2009

### rafehi

Thanks, both of you.

The required answer was x=3. My younger brother asked me for help with his homework and my confidence was a bit bruised when after 20 minutes I couldn't solve the problem.

The question was finding the value of x for which f(x) = inverse of f(x). Actually, I'm not sure I've ever been required to solve a quartic in school/1st year uni maths, though should have figured it wouldn't be dissimilar to solving a cubic. Problem being I never even tried to solve the quartic, because I figured it'd be above what was expected of the students (as opposed to just plugging it into a calculator and looking for the intercepts). I thought I remembered a simpler method from high school - clearly, I was wrong. Might have been confusing it with a differentiation question.

Again, thanks to both for your replies.

5. Feb 25, 2009

### Mentallic

While cubics and quartics are out of range for highschool students, this doesn't mean they don't appear. There are usually enough rational roots and other helpful info to factor it into a quadratic or so.

x=3 is one of the solutions, but look closely at the solutions for the quadratic $$x^2+x-5=0$$ and test them (quickly testing numerical approximations will do) in the original equation.

6. Feb 25, 2009

### HallsofIvy

You are right that -2 does not satisfy the original equation. I was referring to the polynomial equation ravfehi gave and did not look to see if satified the original equation.

If $x^2-6 = \sqrt{x+6}$, then, squaring both sides $x^4- 12x^2+ 36= x+ 6$ or $x^5- 12x^2- x+ 30= 0$ is a necessary condition but since squaring both sides of an equation may introduce new roots, not a sufficient condition. Every root of the square root equation must be a root of the quartic but roots of the quartic are not necessarily roots of the square root equation. 3 is a root of the quartic that does satisfy the square root equation. -2 is a root of the quartic that does NOT satisfy the square root equation. The other two solutions of the quartic involve irrational square roots and do not satisfy the square root equation. (The square root of an irrational square root is algebraic of order 4 while its square is algebraic of order 2- they cannot be equal. But you can also do a numerical check as Mentalic said. 3 is the only root of $x^2- 6= \sqrt{x+ 6}$.