Question

in the given equation, $a$ and $b$ are positive constants. the sum of the solutions to the given equation is $k(4a + b)$, where $k$ is a constant. what is the value of $k$?
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$64x^{2}-(16a + 4b)x+ab = 0$

Explanation:

Step1: Recall the sum - of - roots formula

For a quadratic equation $Ax^{2}+Bx + C=0$, the sum of the roots $x_1 + x_2=-\frac{B}{A}$. In the given equation $64x^{2}-(16a + 4b)x+ab = 0$, we have $A = 64$ and $B=-(16a + 4b)$.

Step2: Calculate the sum of the roots

Using the formula, the sum of the roots $x_1 + x_2=\frac{16a + 4b}{64}$.

Step3: Simplify the expression

$\frac{16a + 4b}{64}=\frac{4(4a + b)}{64}=\frac{4a + b}{16}=\frac{1}{16}(4a + b)$.
Since the sum of the solutions is $k(4a + b)$, then $k=\frac{1}{16}$.

Answer:

$\frac{1}{16}$