How are you doing today, Lightning?
You may know that you can determine the equation of the axis of symmetry of any parabola by knowing that \(x=\frac{-b}{2a}\) . In this case, the axis of symmetry is already given: \(x=-3\) .
Since \(x=-3\) and \(x=\frac{-b}{2a}\), \(\frac{-b}{2a}=-3\) . The only task left is to solve for b/a .
\(\frac{-b}{2a}=-3\) | Multiply by 2 on both sides. |
\(\frac{-b}{a}=-6\) | We are almost there! Multiply by -1 to inverse the sign. |
\(\frac{b}{a}=6\) | There you go! |
In order to answer this question, one must determine what the solutions are to the equation \(18+5x^2=20x\). Let's do that, shall we?
\(18+5x^2=20x\) | When solving quadratic equations, it is generally advisable to move all terms to one side of the equation. |
\(5x^2-20x+18=0\) | This equation is probably best solved with the quadratic formula. |
\(a=5; b=-20; c=18\\ x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) | Now, plug into this equation and find the roots. |
\(x_{1,2}=\frac{-(-20)\pm\sqrt{(-20)^2-4*5*18}}{2*5}\) | The rest is a matter of simplifying accurately. |
\(x_{1,2}=\frac{20\pm\sqrt{40}}{10}\) | Generally, there is a need to put radicals into simplest form, but this is not necessary since the question asks for their rounded value. |
\(x_1\approx3\\ x_2\approx1\) | Let's find the rounded roots' product. |
\(x_1*x_2=3*1=3\) | This is it! |