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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

The solutions to $x^{2} + 6 x + 7 = 0$ are $r$ and $s$ , where $r < s$ . The solutions to $x^{2} + 8 x + 8 = 0$ are $t$ and $u$ , where $t < u$ . The solutions to $x^{2} + 14 x + c = 0$ , where $c$ is a constant, are $r + t$ and $s + u$ . What is the value of $c$ ?

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Explanation

The correct answer is $31$ . Subtracting $7$ from both sides of the equation $x^{2} + 6 x + 7 = 0$ yields $x^{2} + 6 x = - 7$ . To complete the square, adding ${(\frac{6}{2})}^{2}$ , or $3 squared$ , to both sides of this equation yields $x^{2} + 6 x + 3^{2} = - 7 + 3^{2}$ , or ${(x + 3)}^{2} = 2$ . Taking the square root of both sides of this equation yields $x + 3 = \pm \sqrt{2}$ . Subtracting $3$ from both sides of this equation yields $x = - 3 \pm \sqrt{2}$ . Therefore, the solutions $r$ and $s$ to the equation $x^{2} + 6 x + 7 = 0$ are $- 3 - \sqrt{2}$ and $- 3 + \sqrt{2}$ . Since $r < s$ , it follows that $r = - 3 - \sqrt{2}$ and $s = - 3 + \sqrt{2}$ . Subtracting $8$ from both sides of the equation $x^{2} + 8 x + 8 = 0$ yields $x^{2} + 8 x = - 8$ . To complete the square, adding ${(\frac{8}{2})}^{2}$ , or $4 squared$ , to both sides of this equation yields $x^{2} + 8 x + 4^{2} = - 8 + 4^{2}$ , or ${(x + 4)}^{2} = 8$ . Taking the square root of both sides of this equation yields $x + 4 = \pm \sqrt{8}$ , or $x + 4 = \pm 2 \sqrt{2}$ . Subtracting $4$ from both sides of this equation yields $x = - 4 \pm 2 \sqrt{2}$ . Therefore, the solutions $t$ and $u$ to the equation $x^{2} + 8 x + 8 = 0$ are $- 4 - 2 \sqrt{2}$ and $- 4 + 2 \sqrt{2}$ . Since $t < u$ , it follows that $t = - 4 - 2 \sqrt{2}$ and $u = - 4 + 2 \sqrt{2}$ . It's given that the solutions to $x^{2} + 14 x + c = 0$ , where $c$ is a constant, are $r + t$ and $s + u$ . It follows that this equation can be written as $(x - (r + t)) (x - (s + u)) = 0$ , which is equivalent to $x^{2} - (r + t + s + u) x + (r + t) (s + u) = 0$ . Therefore, the value of $c$ is $(r + t) (s + u)$ . Substituting $- 3 - \sqrt{2}$ for $r$ , $- 4 - 2 \sqrt{2}$ for $t$ , $- 3 + \sqrt{2}$ for $s$ , and $- 4 + 2 \sqrt{2}$ for $u$ in this equation yields $((- 3 - \sqrt{2}) + (- 4 - 2 \sqrt{2})) ((- 3 + \sqrt{2}) + (- 4 + 2 \sqrt{2}))$ , which is equivalent to $(- 7 - 3 \sqrt{2}) (- 7 + 3 \sqrt{2})$ , or $(- 7) (- 7) - (3 \sqrt{2}) (3 \sqrt{2})$ , which is equivalent to $49 minus 18$ , or $31$ . Therefore, the value of $c$ is $31$ .