HELP ME

A parabola with equation y=ax^2+bx+c has a vertical line of symmetry at x=2 and goes through the two points (1,1) and (4,-1). The quadratic ax^2 + bx +c has two real roots. The greater root is sqrt(n)+2. What is n?

 

has a vertical line of symmetry at x=2    so     

\(\frac{-b}{2a}=2\\ b=-4a\)

 

\(y=ax^2+bx+c\\ y=ax^2-4ax+c\)

 

goes through the two points (1,1) and (4,-1).

 

\(y=ax^2-4ax+c\\ 1=a-4a+c\\ 1=-3a+c\\ c=1+3a\\~\\ y=ax^2-4ax+1+3a\\ -1=16a-16a+1+3a\\ -2=3a\\ a=-2/3\\~\\ y=\frac{-2x^2}{3}+\frac{8x}{3}-1\\ \)

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = \frac{\frac{-8}{3} \pm \sqrt{\frac{64}{9}-\frac{8}{3}} }{\frac{-4}{3}}\\ x = \frac{\frac{8}{3} \pm \sqrt{\frac{64}{9}-\frac{8}{3}} }{\frac{4}{3}}\\ x = \frac{\frac{8}{3} \pm \sqrt{\frac{64}{9}-\frac{24}{9}} }{\frac{4}{3}}\\ x = \frac{\frac{8}{3} \pm \sqrt{\frac{40}{9}} }{\frac{4}{3}}\\ x = \frac{8 \pm \sqrt{40} }{4}\\ x=2\pm\frac{\sqrt{10}}{2}\\ \text{The great root is}\\ x=2+\frac{\sqrt{10}}{\sqrt4}\\ x=2+\frac{\sqrt{10}}{\sqrt4}\\ x=\sqrt{2.5}+2\)

 

n=2.5