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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
2 years, 1 month ago

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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. Abhinav Raichur · 2 years ago

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@Abhinav Raichur 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. Karthik Venkata · 2 years ago

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@Karthik Venkata 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 \[x=\frac{-b}{2a}\] Abhinav Raichur · 2 years ago

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@Abhinav Raichur Well excellent proof Abhinav Raichur ! Karthik Venkata · 2 years ago

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@Karthik Venkata thank you :) Abhinav Raichur · 2 years ago

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