×

# A simple proof ?

If $$f(x)$$ is a quadratic equation with $$2$$ real roots, prove that the points where the graph of $$y=f(x)$$ cuts the $$x-$$ axis are equidistant to every point on the symmetric axis of $$f(x)$$.

Note by Karthik Venkata
1 year, 11 months ago

Sort by:

consider the function to be $F(x) = ax^2 + bx + c$ given that roots are real .... two cases arise

(1) * they are equal *

i.e they both lie on axis of symmetry and share the same point hence are equidistant.

(2) they are distinct

the formula for roots is given by $x = \frac{-b}{2a} + or - \frac{\sqrt{b^2-4ac}}{2a}$ ......... {1}

which shows that equal amounts of $\frac{\sqrt{b^2-4ac}}{2a}$ is added and subtracted from the axis of symmetry giving rise to the roots. {this tells that the x coordinates of roots are placed equally apart from axis of symmetry}

they lie on the same horizontal axis {implies y coordinate w.r.t point on axis of symmetry is same} hence roots are equidistant. · 1 year, 10 months ago

You are really close to the actual proof, you just need to prove that the quadratic has its symmetric axis given by the plot of the equation $$x = \dfrac{-b}{2a}$$ in the $$xy$$ plane. · 1 year, 10 months ago

oops i considered that to be true ........ we can write F(x) as $F(x) = (x+\frac{b}{2a} )^2 + \frac{b^2 - 4ac}{4 a^2}$ comparing this with the graph of $f(x) = x^2 + k$ which has symmetry at x=0 {the y axis} , after translation of $\frac{b}{2a}$ the symmetry shifts to $x=\frac{-b}{2a}$ · 1 year, 10 months ago

Well excellent proof Abhinav Raichur ! · 1 year, 10 months ago