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 \( f(x) \).

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.

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.

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 \[x=\frac{-b}{2a}\]

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TopNewestconsider 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 distinctthe 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.

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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.

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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 \[x=\frac{-b}{2a}\]Log in to reply

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