what 2 numbers add to get -3 and multiply to get -32?
MathBot Answer:
The numbers are \(- \frac{\sqrt{137}}{2} - \frac{3}{2}\) and \(- \frac{3}{2} + \frac{\sqrt{137}}{2}\).
The sum of the numbers is \(-3\), and their product is \(-32\). This translates to the two equations: \[\begin{aligned}n_{1} + n_{2} &= -3\\n_{1} n_{2} &= -32\\\end{aligned}\] Solving the first equation for \(n_{2}\) in terms of \(n_{1}\) produces the formula \(n_{2} = - n_{1} - 3\).
Substituting \(- n_{1} - 3\) for \(n_{2}\) in the second equation produces a quadratic equation that can be solved by the quadratic formula:\[\begin{gather*}n_{1} \cdot \left( - n_{1} - 3 \right) = -32\\n_{1}^{2} + 3 n_{1} - 32 = 0\end{gather*}\] The solutions to the quadratic equation are \(n_{1} = - \frac{\sqrt{137}}{2} - \frac{3}{2}\) and \(n_{1} = - \frac{3}{2} + \frac{\sqrt{137}}{2}\).
Substituting \(n_{1} = - \frac{\sqrt{137}}{2} - \frac{3}{2}\) into \(n_{1} + n_{2} = -3\), and solving for \(n_{2}\) yields \(n_{2} = - \frac{3}{2} + \frac{\sqrt{137}}{2}\). Similarly, \(n_{1} = - \frac{3}{2} + \frac{\sqrt{137}}{2}\) implies \(n_{2} = - \frac{\sqrt{137}}{2} - \frac{3}{2}\).