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

$64 x^{2} - (16 a + 4 b) x + a b = 0$

In the given equation, $a$ and $b$ are positive constants. The sum of the solutions to the given equation is $k left parenthesis 4 a plus b right parenthesis$ , where $k$ is a constant. What is the value of $k$ ?

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Explanation

The correct answer is $\frac{1}{16}$ . Let $p$ and $q$ represent the solutions to the given equation. Then, the given equation can be rewritten as $64 (x - p) (x - q) = 0$ , or $64 x^{2} - 64 (p + q) + p q = 0$ . Since this equation is equivalent to the given equation, it follows that $- (16 a + 4 b) = - 64 (p + q)$ . Dividing both sides of this equation by $negative 64$ yields $\frac{16 a + 4 b}{64} = p + q$ , or $\frac{1}{16} (4 a + b) = p + q$ . Therefore, the sum of the solutions to the given equation, $p + q$ , is equal to $one sixteenth left parenthesis 4 a plus b right parenthesis$ . Since it's given that the sum of the solutions to the given equation is $k left parenthesis 4 a plus b right parenthesis$ , where $k$ is a constant, it follows that $k = \frac{1}{16}$ . Note that 1/16, .0625, 0.062, and 0.063 are examples of ways to enter a correct answer.

Alternate approach: The given equation can be rewritten as $64 x^{2} - 4 (4 a + b) x + a b = 0$ , where $a$ and $b$ are positive constants. Dividing both sides of this equation by $4$ yields $16 x^{2} - (4 a + b) x + \frac{a b}{4} = 0$ . The solutions for a quadratic equation in the form $A x^{2} + B x + C = 0$ , where $A$ , $B$ , and $C$ are constants, can be calculated using the quadratic formula, $x = \frac{- B + \sqrt{B^{2} - 4 A C}}{2 A}$ and $x = \frac{- B - \sqrt{B^{2} - 4 A C}}{2 A}$ . It follows that the sum of the solutions to a quadratic equation in the form $A x^{2} + B x + C = 0$ is $\frac{- B + \sqrt{B^{2} - 4 A C}}{2 A} + \frac{- B - \sqrt{B^{2} - 4 A C}}{2 A}$ , which can be rewritten as $\frac{- B + - B + \sqrt{B^{2} - 4 A C} - \sqrt{B^{2} - 4 A C}}{2 A}$ , which is equivalent to $\frac{- 2 B}{2 A}$ , or $- \frac{B}{A}$ . In the equation $16 x^{2} - (4 a + b) x + \frac{a b}{4} = 0$ , $upper A equals 16$ , $B = - (4 a + b)$ , and $C = \frac{a b}{4}$ . Substituting $16$ for $A$ and $- (4 a + b)$ for $B$ in $- \frac{B}{A}$ yields $- \frac{- (4 a + b)}{16}$ , which can be rewritten as $one sixteenth left parenthesis 4 a plus b right parenthesis$ . Thus, the sum of the solutions to the given equation is $one sixteenth left parenthesis 4 a plus b right parenthesis$ . Since it's given that the sum of the solutions to the given equation is $k left parenthesis 4 a plus b right parenthesis$ , where $k$ is a constant, it follows that $k = \frac{1}{16}$ .