Find all values of x satisfying the given conditions.
-$y _ { 1 } = \frac { 1 } { x + 15 } , y _ { 2 } = \frac { 1 } { x }$, and $y _ { 1 } + y _ { 2 } = \frac { 1 } { 5 }$
A) $\frac { - 5 \pm 5 \sqrt { 13 } } { 2 }$
B) $\frac { - 25 \pm 5 \sqrt { 13 } } { 2 }$
C) $\frac { 25 \pm 5 \sqrt { 13 } } { 2 }$
D) $\frac { 5 \pm 5 \sqrt { 13 } } { 2 }$

Question

Quizplus · Accepted Answer

To solve for $x$, we first combine the expressions for $y_1$ and $y_2$ as given by the equation $y_1 + y_2 = \frac{1}{5}$. Substituting the given expressions for $y_1$ and $y_2$, we get $\frac{1}{x + 15} + \frac{1}{x} = \frac{1}{5}$. To solve this equation, we find a common denominator and simplify:$$\frac{x + (x + 15)}{x(x + 15)} = \frac{1}{5}$$$$\frac{2x + 15}{x^2 + 15x} = \frac{1}{5}$$Multiplying both sides by $5(x^2 + 15x)$ to clear the fraction:$$5(2x + 15) = x^2 + 15x$$Expanding and simplifying:$$10x + 75 = x^2 + 15x$$Rearranging to form a quadratic equation:$$x^2 + 5x - 75 = 0$$Using the quadratic formula, where $a = 1$, $b = 5$, and $c = -75$:$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 + 300}}{2} = \frac{-5 \pm \sqrt{325}}{2} = \frac{-5 \pm 5\sqrt{13}}{2}$$Therefore, the correct answer is $\frac{-5 \pm 5\sqrt{13}}{2}$.

Find All Values of X Satisfying the Given Conditions y1=1x+15,y2=1xy _ { 1 } = \frac { 1 } { x + 15 } , y _ { 2 } = \frac { 1 } { x }y1​=x+151​,y2​=x1​

Find All Values of X Satisfying the Given Conditions $y _ { 1 } = \frac { 1 } { x + 15 } , y _ { 2 } = \frac { 1 } { x }$