a**2+b**2=13

ab=6

a+b=?

asked by guest
on Nov 29, 2024 at 9:29 pm



You asked:

Given the following equations: \[\begin{aligned}{a}^{2} + {b}^{2} &= 13\\ a b &= 6\end{aligned}\] Evaluate \(a + b\).

MathBot Answer:

The system of equations has \(4\) solutions.

\[a = -3, b = -2\]\[a = -2, b = -3\]\[a = 2, b = 3\]\[a = 3, b = 2\]

By substituting these values into the expression, we find that: $$\begin{aligned}a + b &= -5\end{aligned}$$$$\begin{aligned}a + b &= 5\end{aligned}$$

Solve \(a^{2} + b^{2} = 13\) for \(a\). \[a = - \sqrt{13 - b^{2}}, a = \sqrt{13 - b^{2}}\]Substitute \(- \sqrt{13 - b^{2}}\) for \(a\) in \(a b = 6\) and simplify. $$\begin{aligned}a b &amp= 6 \\ \left(- \sqrt{13 - b^{2}}\right) b &= 6 \\ b \sqrt{13 - b^{2}} &= -6 \end{aligned}$$Substitute \(-3\) into \(a^{2} + b^{2} = 13\) to solve for \(a\). $$\begin{aligned}a^{2} + \left(-3\right)^{2} &= 13 \\ a^{2} + 9 &= 13 \\a^{2} - 4 &= 0 \\ \left(a - 2\right) \left(a + 2\right) &= 0 \\ a = -2&, a = 2\end{aligned}$$This yields the following solution. $$\begin{aligned}a = -2,\,b = -3\end{aligned}$$Substitute \(-2\) into \(a^{2} + b^{2} = 13\) to solve for \(a\). $$\begin{aligned}a^{2} + \left(-2\right)^{2} &= 13 \\ a^{2} + 4 &= 13 \\a^{2} - 9 &= 0 \\ \left(a - 3\right) \left(a + 3\right) &= 0 \\ a = -3&, a = 3\end{aligned}$$This yields the following solution. $$\begin{aligned}a = -3,\,b = -2\end{aligned}$$Substitute \(- \sqrt{13 - b^{2}}\) for \(a\) in \(a b = 6\) and simplify. $$\begin{aligned}a b &amp= 6 \\ \left(- \sqrt{13 - b^{2}}\right) b &= 6 \\ b \sqrt{13 - b^{2}} &= -6 \end{aligned}$$Substitute \(\sqrt{13 - b^{2}}\) for \(a\) in \(a b = 6\) and simplify. $$\begin{aligned}a b &amp= 6 \\ \left(\sqrt{13 - b^{2}}\right) b &= 6 \\ b \sqrt{13 - b^{2}} &= 6 \end{aligned}$$Substitute \(2\) into \(a^{2} + b^{2} = 13\) to solve for \(a\). $$\begin{aligned}a^{2} + 2^{2} &= 13 \\ a^{2} + 4 &= 13 \\a^{2} - 9 &= 0 \\ \left(a - 3\right) \left(a + 3\right) &= 0 \\ a = -3&, a = 3\end{aligned}$$This yields the following solution. $$\begin{aligned}a = 3,\,b = 2\end{aligned}$$Substitute \(3\) into \(a^{2} + b^{2} = 13\) to solve for \(a\). $$\begin{aligned}a^{2} + 3^{2} &= 13 \\ a^{2} + 9 &= 13 \\a^{2} - 4 &= 0 \\ \left(a - 2\right) \left(a + 2\right) &= 0 \\ a = -2&, a = 2\end{aligned}$$This yields the following solution. $$\begin{aligned}a = 2,\,b = 3\end{aligned}$$

By substituting these values into the expression, we find that: $$\begin{aligned}a + b &= -5\end{aligned}$$$$\begin{aligned}a + b &= 5\end{aligned}$$

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