Question

determine the equation of the parabola with roots 3+sqrt(5) and 3-sqrt(5), and passing through the point (3/10)

Answer 1

`\begin{cases} x = 3+√(5)\implies &x-3-√(5)=0\\ x= 3-√(5)\implies &x-3+√(5)=0 \end{cases}\implies a(x-3-√(5))(x-3+√(5))=\stackrel{y}{0} \\\\[-0.35em] ~\dotfill\\\\ \underset{\textit{difference of squares}}{[(x-3)~-~(√(5))][(x-3)~+~(√(5))]}\implies (x-3)^2-(√(5))^2 \\\\\\ (x^2-6x+9)-5\implies x^2-6x+4 \\\\[-0.35em] ~\dotfill`

`a(x^2-6x+4)=y\qquad \stackrel{\textit{we also know that}}{x =3 \qquad y = 10}\qquad \implies a[(3)^2-6(3)+4]=10 \\\\\\ a(9-18+4)=10\implies -5a=10\implies a = \cfrac{10}{-5}\implies \underline{a = -2} \\\\\\ -2(x^2-6x+4)=y\implies \boxed{-2x^2+12x-8=y}`

Check the picture below.