If a, b, c are real numbers and ab = 3, bc = 4, and ac = 6, what is 2a2 + 3b2 + c2?

1. Break Down the Problem

We have three equations based on the products of variables:

• $ab = 3$
• $bc = 4$
• $ac = 6$

We need to find the value of $2a^2 + 3b^2 + c^2$.

2. Relevant Concepts

From the given equations, we can derive individual values for $a$, $b$, and $c$ in terms of each other. We will utilize these relationships and the formulas for computing squares.

3. Analysis and Detail

To find $a, b, c$ individually, we can manipulate the equations:

1. From $ab = 3$, we have: $b = \frac{3}{a}$

2. Substitute $b$ into $bc = 4$: $\left(\frac{3}{a}\right)c = 4 \implies c = \frac{4a}{3}$

3. Substitute $b$ and $c$ into $ac = 6$: $a\left(\frac{4a}{3}\right) = 6 \implies \frac{4a^2}{3} = 6 \implies 4a^2 = 18 \implies a^2 = \frac{9}{2} \implies a = \pm \frac{3}{\sqrt{2}} = \pm \frac{3\sqrt{2}}{2}$

4. Now, substitute $a = \frac{3\sqrt{2}}{2}$ into $b = \frac{3}{a}$: $b = \frac{3}{\frac{3\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

5. Substitute $a$ into $c = \frac{4a}{3}$: $c = \frac{4 \cdot \frac{3\sqrt{2}}{2}}{3} = 2\sqrt{2}$

Now we have:

• $a = \frac{3\sqrt{2}}{2}$
• $b = \sqrt{2}$
• $c = 2\sqrt{2}$

4. Verify and Summarize

Now, we compute $2a^2 + 3b^2 + c^2$:

• $a^2 = \left(\frac{3\sqrt{2}}{2}\right)^2 = \frac{9 \cdot 2}{4} = \frac{18}{4} = \frac{9}{2}$
• $b^2 = (\sqrt{2})^2 = 2$
• $c^2 = (2\sqrt{2})^2 = 4 \cdot 2 = 8$

Now, substituting these values back: $2a^2 = 2 \left(\frac{9}{2}\right) = 9$ $3b^2 = 3 \cdot 2 = 6$ $c^2 = 8$

Thus: $2a^2 + 3b^2 + c^2 = 9 + 6 + 8 = 23$

Final Answer

The value of $2a^2 + 3b^2 + c^2$ is $23$.