Positive solutions to $x^2 - 2x - 3 = \sqrt{x+3}$

Question

I found this interesting problem in one of my practice tests

The positive real solution of $x^2-2x-3=\sqrt{x+3}$ has the form of $\frac{a+\sqrt{b}}{c}$ where $a$, $b$, $c$ are primes. Find $a+b+c$.

First of all, I tried to find the range of solutions.

Since $\sqrt{x+3} \ge 0$, that means $x \in [-3, \infty)$

And it also means that $x^2 - 2x - 3 \ge 0$, which means $x \in (-\infty, -1] \cup [3, \infty)$.

Combining the two conditions along with the fact that we're finding positive $x$, we have $x \in [3, \infty)$.

And at this point I don't really know how to continue. There's no substitution that I know that can deal with the square root, squaring both sides and turning it into a quartic equation is definitely not the way here, guess and check is not ideal either. My question is, is there a rigorous way to solve this?

Answer 1

HINT

Notice that $x = 3$ is a root for the expression on the LHS. Consequently, we can factor it as next: \begin{align*} x^{2} - 2x - 3 & = (x^{2} - 3x) + (x - 3)\\\\ & = x(x - 3) + (x - 3)\\\\ & = (x + 1)(x - 3) \end{align*}

Based on the assumption that $x\in[3,+\infty)$, it is possible to go even further: \begin{align*} x^{2} - 2x - 3 = \sqrt{x + 3} & \Longleftrightarrow (x + 1)(x - 3) = \sqrt{x + 3}\\\\ & \Longleftrightarrow ((x + 3) - 2)((x + 3) - 6) = \sqrt{x + 3}\\\\ & \Longleftrightarrow (u^{2} - 2)(u^{2} - 6) = u\\\\ & \Longleftrightarrow u^{4} - 8u^{2} - u + 12 = 0 \end{align*}

Can you take it from here?

Answer 2

It's easy if you can find out this factor, assume that you found the domain is $x\ge3$.

$x^2-2x-3=\sqrt{x+3}$ i.e $x^2-x+\frac{1}{4}=x+3+\sqrt{x+3}+\frac{1}{4}$ i.e

$(x-\frac{1}{2})^2=(\sqrt{x+3}+\frac{1}{2})^2$

Now, your equation became easier, and for each case, raise both sides by power of 2 only leads to quadratic equation which is trivial to solve.

Answer 3

The equation $ \ x^2 - 2x - 3 \ = \ \sqrt{x+3} \ $ can be viewed as describing the intersections of the "vertical" parabola $ \ f(x) \ = \ x^2 - 2x - 3 \ $ with the "upper half" of the "horizontal" parabola $ \ y^2 \ = \ x + 3 \ \ . \ $ As you've already found from the domains and ranges, one intersection point must lie in the interval $ \ (-3 \ , -1) \ $ and the other in $ \ (3 \ , \ \infty) \ \ . $ The intermediate value theorem allows us to refine this a bit, since $ \ f(-2) \ = \ f(+4) \ = \ 5 \ $ and $ \ f(-1) \ = \ f(+3) \ = \ 0 \ \ , \ $ while $ \ 1 \ < \ g(x) \ = \ \sqrt{x+3} \ < \ \sqrt7 \ $ on $ \ (-2 \ , \ 4) \ \ . \ $ The two intersections are thus found in $ \ (-2 \ , -1) \ $ and the other in $ \ (3 \ , \ 4) \ \ . $

Since $ \ f(x) \ = \ (x - 1)^2 - 4 \ \ , \ $ we can separate its parabola into the two "halves" about its symmetry axis, $ \ x \ = \ 1 \ \pm \ \sqrt{y + 4} \ \ . $ The "right half" $ \ x \ = \ 1 \ + \ \sqrt{y + 4} \ $ is the portion [in blue in the graph below] that intersects $ \ y \ = \ \sqrt{x + 3} \ $ [in red] in the interval $ \ (3 \ , \ 4) \ \ . \ $ This parabola has its vertex at $ \ (1 \ , \ -4) \ $ and $ \ y^2 \ = \ x + 3 \ $ has its vertex at $ \ (-3 \ , \ 0) \ \ , \ $ and both quadratic polynomials have the same quadratic coefficient, these two square-root curves have the same "shape", so they are simply translated and reflected versions of one another. The line connecting their vertices has slope $ \ \frac{-4 \ - \ 0}{1 \ - \ (-3)} \ = \ -1 \ \ , \ $ therefore the two square-root curves are reflections of each other about the perpendicular bisector of the segment connecting their vertices, $ \ y - (-2) \ = \ +1·( \ x - [-1] \ ) $ $ \rightarrow \ y \ = \ x - 1 \ \ $ [marked in green].

The two square-root curves therefore intersect on this line (at the fixed-point of the transformation $ \ y \ = \ x - 1 \ ^{*} \ ) \ , $ permitting us to solve for this intersection using either $$ \sqrt{x + 3} \ \ = \ \ x \ - \ 1 \ \ \Rightarrow \ \ x \ + \ 3 \ \ = \ \ x^2 \ - \ 2x \ + \ 1 \ \ \Rightarrow \ \ x^2 \ - \ 3x \ - \ 2 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ x \ \ = \ \ \frac{3 \ \pm \ \sqrt{17}}{2} $$ or $$ 1 \ + \ \sqrt{y + 4} \ \ = \ \ y \ + \ 1 \ \ \Rightarrow \ \ y^2 \ - \ y \ - \ 4 \ \ = \ \ 0 \ \ \Rightarrow \ \ y \ \ = \ \ \frac{1 \ \pm \ \sqrt{17}}{2} \ \ . $$

$ \ ^{*} $ We see that this is the applicable transformation from $ \ x \ = \ 1 + \sqrt{y + 4} $ $ \rightarrow \ \ (y + 1) \ = \ 1 + \sqrt{(x - 1) + 4} \ \ \rightarrow \ \ y \ = \ \sqrt{ x + 3} \ \ . $

Since $ \ \frac{3 \ + \ \sqrt{17}}{2} \ \approx \ 3.562 \ $ lies within $ \ (3 \ , \ 4) \ \ , \ $ this is the positive real solution we seek for the original equation (whereas $ \ \frac{3 \ - \ \sqrt{17}}{2} \ < \ 0 \ $ is "spurious"); $ \ y \ \ = \ \ \frac{1 \ + \ \sqrt{17}}{2} \ = \ x - 1 \ $ is consistent with this result. We conclude that $ \ \mathbf{a + b + c \ = \ 3 + 17 + 2 \ = \ 22} \ \ . $

The intersection point in $ \ (-2 \ , \ -1) \ $ lies on the line $ \ y \ = \ -x \ \ $ (which gives us $ \ x \ = \ \frac{1 \ - \ \sqrt{13}}{2} \ ) \ , \ $ but it is not as straightforward to show this, as it is not a line of reflection for the curves.