Question

one of the solutions to $x^{2}-2x-15=0$ is $x=-3$. what is the other solution?
$x=-5$
$x=-1$
$x=1$
$x=5$

Explanation:

Step1: Use sum of roots property

For quadratic $ax^2+bx+c=0$, sum of roots is $-\frac{b}{a}$.
Here, $a=1$, $b=-2$, so sum of roots = $-\frac{-2}{1}=2$.

Step2: Calculate the other root

Let the other root be $r$. We know one root is $-3$, so:
$r + (-3) = 2$
$r = 2 + 3$

Answer:

D. $x = 5$