solve for x in the equation $x^{2}+20x + 100 = 36$.\n$x=-16$ or $x = - 4$\n$x=-10$\n$x=-8$\n$x = 4$ or $x =…

solve for x in the equation $x^{2}+20x + 100 = 36$.\n$x=-16$ or $x = - 4$\n$x=-10$\n$x=-8$\n$x = 4$ or $x = 16$

solve for x in the equation $x^{2}+20x + 100 = 36$.\n$x=-16$ or $x = - 4$\n$x=-10$\n$x=-8$\n$x = 4$ or $x = 16$

Answer

Explanation:

Step1: Rewrite the left - hand side as a perfect square.

The left - hand side of the equation $x^{2}+20x + 100$ is a perfect square trinomial. Using the formula $(a + b)^2=a^{2}+2ab + b^{2}$, where $a = x$ and $b = 10$, we can rewrite the equation as $(x + 10)^{2}=36$.

Step2: Take the square root of both sides.

Taking the square root of both sides gives $x+10=\pm\sqrt{36}$, so $x + 10=\pm6$.

Step3: Solve for x in two cases.

Case 1: When $x + 10=6$, then $x=6 - 10=-4$. Case 2: When $x + 10=-6$, then $x=-6 - 10=-16$.

Answer:

$x=-16$ or $x=-4$