Question

what values of b satisfy 3(2b + 3)² = 36?
○ b = (-3 + 2√3)/2 and (-3 - 2√3)/2
○ b = (-3 + 2√3)/3 and (-3 - 2√3)/3
○ b = 3/2 and -9/2
○ b = 9/2 and -3/2

Explanation:

Step1: Divide both sides by 3

To isolate the squared term, divide each side of the equation \(3(2b + 3)^2 = 36\) by 3.
\(\frac{3(2b + 3)^2}{3} = \frac{36}{3}\)
Simplifies to: \((2b + 3)^2 = 12\)

Step2: Take square root of both sides

Take the square root of both sides to eliminate the square. Remember to consider both positive and negative roots.
\(2b + 3 = \pm\sqrt{12}\)
Simplify \(\sqrt{12}\) to \(2\sqrt{3}\), so: \(2b + 3 = \pm 2\sqrt{3}\)

Step3: Solve for \(b\)

Subtract 3 from both sides: \(2b = -3 \pm 2\sqrt{3}\)
Then divide both sides by 2: \(b = \frac{-3 \pm 2\sqrt{3}}{2}\)

Wait, hold on—wait, no, wait. Wait, in Step 1, we had \((2b + 3)^2 = 12\), then square root: \(2b + 3 = \pm \sqrt{12} = \pm 2\sqrt{3}\). Then subtract 3: \(2b = -3 \pm 2\sqrt{3}\), then divide by 2: \(b = \frac{-3 \pm 2\sqrt{3}}{2}\). Wait, but let's check the answer options. Wait, the first option is \(b = \frac{-3 + 2\sqrt{3}}{2}\) and \(\frac{-3 - 2\sqrt{3}}{2}\). Wait, but let's check the options again. Wait, the third option is \(b = \frac{3}{2}\) and \(-\frac{9}{2}\). Wait, maybe I made a mistake. Wait, no—wait, wait, let's re-express. Wait, maybe I miscalculated. Wait, let's start over.

Wait, original equation: \(3(2b + 3)^2 = 36\). Divide both sides by 3: \((2b + 3)^2 = 12\). Take square roots: \(2b + 3 = \pm \sqrt{12} = \pm 2\sqrt{3}\). Then, \(2b = -3 \pm 2\sqrt{3}\), so \(b = \frac{-3 \pm 2\sqrt{3}}{2}\). So \(b = \frac{-3 + 2\sqrt{3}}{2}\) and \(b = \frac{-3 - 2\sqrt{3}}{2}\), which is the first option. Wait, but let's check the third option: \(b = \frac{3}{2}\) and \(-\frac{9}{2}\). Let's test \(b = \frac{3}{2}\): plug into original equation. \(3(2*(\frac{3}{2}) + 3)^2 = 3(3 + 3)^2 = 3(6)^2 = 3*36 = 108 eq 36\). So that's wrong. Wait, but wait, maybe I made a mistake in the square root. Wait, no—wait, \(\sqrt{12}\) is \(2\sqrt{3}\), correct. So the first option is \(b = \frac{-3 + 2\sqrt{3}}{2}\) and \(\frac{-3 - 2\sqrt{3}}{2}\), which matches the first option. Wait, but let's check the third option. Let's plug \(b = \frac{3}{2}\) into the original equation: \(3(2*(\frac{3}{2}) + 3)^2 = 3(3 + 3)^2 = 3*36 = 108 eq 36\). So that's incorrect. Wait, but the first option's expressions are \(\frac{-3 + 2\sqrt{3}}{2}\) and \(\frac{-3 - 2\sqrt{3}}{2}\). Let's verify with \(b = \frac{-3 + 2\sqrt{3}}{2}\):

Compute \(2b + 3 = 2*(\frac{-3 + 2\sqrt{3}}{2}) + 3 = -3 + 2\sqrt{3} + 3 = 2\sqrt{3}\). Then square that: \((2\sqrt{3})^2 = 12\). Multiply by 3: \(3*12 = 36\), which matches. Similarly for the negative root: \(2b + 3 = 2*(\frac{-3 - 2\sqrt{3}}{2}) + 3 = -3 - 2\sqrt{3} + 3 = -2\sqrt{3}\). Square that: \((-2\sqrt{3})^2 = 12\). Multiply by 3: \(3*12 = 36\), which also matches. So the first option is correct. Wait, but let's check the options again. The first option is:

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

Yes, that's what we derived. So the correct answer is the first option. Wait, but let's check the third option. Wait, maybe I made a mistake in the problem. Wait, no—wait, the third option is \(b = \frac{3}{2}\) and \(-\frac{9}{2}\). Let's solve the equation another way. Let's expand the original equation: \(3(4b^2 + 12b + 9) = 36\) → \(12b^2 + 36b + 27 = 36\) → \(12b^2 + 36b - 9 = 0\) → Divide by 3: \(4b^2 + 12b - 3 = 0\). Use quadratic formula: \(b = \frac{-12 \pm \sqrt{144 + 48}}{8} = \frac{-12 \pm \sqrt{192}}{8} = \frac{-12 \pm 8\sqrt{3}}{8} = \frac{-3 \pm 2\sqrt{3}}{2}\), which matches the first option. So the first option is correct. Wait, but the third option: let's check \(b = \frac{3}{2}\): plug into \(4b^2 + 12b - 3\): \(4*(9/4) + 12*(3/2) - 3 = 9 + 18 - 3 = 24 ≠ 0\). So incorrect. So the first option is correct.

Answer:

A. \( b = \frac{-3 + 2\sqrt{3}}{2} \) and \( \frac{-3 - 2\sqrt{3}}{2} \)