Algebra Workout #2

$\begin{aligned} x^2+20x+29 & = 2 \sqrt{x^2+20x+64} \\ \color{#3D99F6}{x^2+20x+64} - 35 & = 2 \sqrt{\color{#3D99F6}{x^2+20x+64}} \quad \quad \small \color{#3D99F6}{\text{Let } y = \sqrt{x^2+20x+64} } \\ \color{#3D99F6}{y}^2 - 2\color{#3D99F6}{y} - 35 & = 0 \\ \Rightarrow (y-7)(y+5) & = 0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Since } y = \sqrt{x^2+20x+64} > 0} \\ \Rightarrow y & = 7 \\ \Rightarrow x^2+20x+64 & = 7^2 \\ x^2+20x+\color{#3D99F6}{15} & = 0 \end{aligned}$

By Vieta's formula, the product of the solutions of $x$ is $\boxed{\color{#3D99F6}{15}}$ .

This problem requires a lot of creativity:

First, we can see that $x^{2} + 20x + 64 = (x+16)(x+4).$ Then, we can express $x^{2} + 20x + 29$ as $(x+16)(x+4)-35.$ Now, we can express $(x+16)(x+4)$ as y. Now we have the simple equation, $y-35 = 2\sqrt{y}.$ From here, we can either factor this into a binomial, or just see that y = 49. When factoring into a binomial, take notice that x= 25 is not a solution because that makes $y-35$ equal -5 and $2\sqrt{y}$ equal 5. With this, we can substitute $(x+16)(x+4)$ into y to get $(x+16)(x+4)$ = 49. Expanding gives $x^{2} + 20x + 15.$ Completing the square gives $(x+10)^{2} -85 = 0$ , which gives $x = \pm\sqrt{85} -10$ The product of the two solutions are difference of squares: $-\sqrt{85}^2 + 10^{2} = 15$ .