Math Problem Solver

\(3 - 3 \sqrt{2}\), \(- 3 \sqrt{2} - 3\) and \(3 + 3 \sqrt{2}\), \(-3 + 3 \sqrt{2}\) are the numbers whose difference is \(6\), and whose product is \(9\).

The difference between the numbers is \(6\), and their product is \(9\). This translates to the two equations: \[\begin{aligned}n_{1} - n_{2} &= 6\\n_{1} n_{2} &= 9\\\end{aligned}\] Solving the first equation for \(n_{2}\) in terms of \(n_{1}\) produces the formula \(n_{2} = n_{1} - 6\).
Substituting \(n_{1} - 6\) for \(n_{2}\) in the second equation produces a quadratic equation that can be solved by the quadratic formula:\[\begin{gather*}n_{1} \cdot \left( n_{1} - 6 \right) = 9\\n_{1}^{2} - 6 n_{1} - 9 = 0\end{gather*}\] The solutions to the quadratic equation are \(n_{1} = 3 - 3 \sqrt{2}\) and \(n_{1} = 3 + 3 \sqrt{2}\).
Substituting \(n_{1} = 3 - 3 \sqrt{2}\) into \(n_{1} - n_{2} = 6\), and solving for \(n_{2}\) yields \(n_{2} = - 3 \sqrt{2} - 3\). Similarly, \(n_{1} = 3 + 3 \sqrt{2}\) implies \(n_{2} = -3 + 3 \sqrt{2}\).