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

$x (x + 1) - 56 = 4 x (x - 7)$

What is the sum of the solutions to the given equation?

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Explanation

The correct answer is $StartFraction 29 Over 3 EndFraction$ . Applying the distributive property to the left-hand side of the given equation, $x (x + 1) - 56$ , yields $x^{2} + x - 56$ . Applying the distributive property to the right-hand side of the given equation, $4 x left parenthesis x minus 7 right parenthesis$ , yields $4 x squared minus 28 x$ . Thus, the equation becomes $x^{2} + x - 56 = 4 x^{2} - 28 x$ . Combining like terms on the left- and right-hand sides of this equation yields $0 = (4 x^{2} - x^{2}) + (- 28 x - x) + 56$ , or $3 x^{2} - 29 x + 56 = 0$ . For a quadratic equation in the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants, the quadratic formula gives the solutions to the equation in the form $x = \frac{(- b \pm \sqrt{b^{2} - 4 a c})}{2 a}$ . Substituting $3$ for $a$ , $negative 29$ for $b$ , and $56$ for $c$ from the equation $3 x^{2} - 29 x + 56 = 0$ into the quadratic formula yields $x = \frac{(29 \pm \sqrt{{(- 29)}^{2} - 4 (3) (56)})}{2 (3)}$ , or $x = \frac{29}{6} \pm \frac{13}{6}$ . It follows that the solutions to the given equation are $\frac{29}{6} + \frac{13}{6}$ and $\frac{29}{6} - \frac{13}{6}$ . Adding these two solutions gives the sum of the solutions: $\frac{29}{6} + \frac{13}{6} + \frac{29}{6} - \frac{13}{6}$ , which is equivalent to $\frac{29}{6} + \frac{29}{6}$ , or $StartFraction 29 Over 3 EndFraction$ . Note that 29/3, 9.666, and 9.667 are examples of ways to enter a correct answer.