what values of b satisfy 3(2b + 3)^2 = 36?\no b = (-3 + 2\\sqrt{3})/2 and (-3 - 2\\sqrt{3})/2\no b = (-3 +…

what values of b satisfy 3(2b + 3)^2 = 36?\no b = (-3 + 2\\sqrt{3})/2 and (-3 - 2\\sqrt{3})/2\no b = (-3 + 2\\sqrt{3})/3 and (-3 - 2\\sqrt{3})/3\no b = 3/2 and -9/2\no b = 9/2 and -3/2

what values of b satisfy 3(2b + 3)^2 = 36?\no b = (-3 + 2\\sqrt{3})/2 and (-3 - 2\\sqrt{3})/2\no b = (-3 + 2\\sqrt{3})/3 and (-3 - 2\\sqrt{3})/3\no b = 3/2 and -9/2\no b = 9/2 and -3/2

Answer

Explanation:

Step1: Divide both sides by 3

Divide the equation $3(2b + 3)^2=36$ by 3 to get $(2b + 3)^2 = 12$.

Step2: Take square - root of both sides

Taking the square - root of both sides gives $2b+3=\pm\sqrt{12}=\pm2\sqrt{3}$.

Step3: Solve for b when $2b + 3=2\sqrt{3}$

Subtract 3 from both sides: $2b=2\sqrt{3}-3$. Then divide by 2: $b=\frac{-3 + 2\sqrt{3}}{2}$.

Step4: Solve for b when $2b + 3=-2\sqrt{3}$

Subtract 3 from both sides: $2b=-2\sqrt{3}-3$. Then divide by 2: $b=\frac{-3 - 2\sqrt{3}}{2}$.

Answer:

$b=\frac{-3 + 2\sqrt{3}}{2}$ and $b=\frac{-3 - 2\sqrt{3}}{2}$