solve $x^{2}+14x = - 24$ by completing the square. what is the solution set of the equation?\n{-12…

solve $x^{2}+14x = - 24$ by completing the square. what is the solution set of the equation?\n{-12, -2}\n{-7, 7}\n{-6, -4}\n{-5, 5}

solve $x^{2}+14x = - 24$ by completing the square. what is the solution set of the equation?\n{-12, -2}\n{-7, 7}\n{-6, -4}\n{-5, 5}

Answer

Explanation:

Step1: Add the square of half - coefficient of x.

The coefficient of $x$ is $14$. Half of it is $7$, and its square is $49$. Add $49$ to both sides of the equation $x^{2}+14x=-24$. So we get $x^{2}+14x + 49=-24 + 49$.

Step2: Rewrite the left - hand side as a perfect square.

The left - hand side $x^{2}+14x + 49=(x + 7)^{2}$, and the right - hand side $-24 + 49 = 25$. So the equation becomes $(x + 7)^{2}=25$.

Step3: Take the square root of both sides.

We have $x+7=\pm\sqrt{25}=\pm5$.

Step4: Solve for x.

When $x + 7 = 5$, then $x=5 - 7=-2$. When $x + 7=-5$, then $x=-5 - 7=-12$.

Answer:

{-12, -2}