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

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

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

Answer

Explanation:

Step1: Factor the quadratic equation

For the quadratic equation $x^{2}-2x - 15=0$, we need to find two numbers that multiply to - 15 and add up to - 2. The numbers are - 5 and 3. So we can factor the equation as $(x + 3)(x - 5)=0$.

Step2: Use the zero - product property

The zero - product property states that if $ab = 0$, then either $a=0$ or $b = 0$. Since $(x + 3)(x - 5)=0$, we have $x+3 = 0$ or $x - 5=0$. We already know that $x=-3$ is a solution from $x + 3=0$. Solving $x - 5=0$ gives $x = 5$.

Answer:

D. $x = 5$