theo started to solve the quadratic equation $(x + 2)^2-9=-5$. he added 9 to both sides and the resulting…

theo started to solve the quadratic equation $(x + 2)^2-9=-5$. he added 9 to both sides and the resulting equation was $(x + 2)^2 = 4$. next, he took the square root of each side. which was the resulting equation of that step?\n$x + 2=pm4$\n$x + 2=pm2$\n$x + 2=pm8$\n$x + 2=pm16$

theo started to solve the quadratic equation $(x + 2)^2-9=-5$. he added 9 to both sides and the resulting equation was $(x + 2)^2 = 4$. next, he took the square root of each side. which was the resulting equation of that step?\n$x + 2=pm4$\n$x + 2=pm2$\n$x + 2=pm8$\n$x + 2=pm16$

Answer

Explanation:

Step1: Recall square - root property

If (y^{2}=a) ((a\geq0)), then (y = \pm\sqrt{a}). Here (y=x + 2) and (a = 4).

Step2: Take square - root of both sides

Given ((x + 2)^{2}=4), we have (x + 2=\pm\sqrt{4}). Since (\sqrt{4}=2), the equation becomes (x + 2=\pm2).

Answer:

B. (x + 2=\pm2)