solve $x^{2}+6x = 7$ by completing the square. which is the solution set of the equation?\n{-7, 1}\n{-3…

Explanation:

Step1: Rewrite the equation

Given $x^{2}+6x = 7$. Add the square of half of the coefficient of $x$ to both sides. The coefficient of $x$ is $6$, half of it is $3$ and its square is $9$. So we have $x^{2}+6x + 9=7 + 9$.

Step2: Factor the left - hand side

The left - hand side $x^{2}+6x + 9$ can be factored as $(x + 3)^{2}$ according to the perfect - square formula $(a + b)^{2}=a^{2}+2ab + b^{2}$ (here $a=x$ and $b = 3$). So $(x + 3)^{2}=16$.

Step3: Solve for $x$

Take the square root of both sides: $x+3=\pm\sqrt{16}=\pm4$. When $x + 3=4$, then $x=4 - 3=1$. When $x + 3=-4$, then $x=-4 - 3=-7$.

Answer:

A. ${-7,1}$