A simple proof ?

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

#Proofs #QuadraticPolynomial

Note by Venkata Karthik Bandaru
6 years, 1 month ago

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Comments

consider the function to be F(x)=ax2+bx+cF(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=b2a+orb24ac2a x = \frac{-b}{2a} + or - \frac{\sqrt{b^2-4ac}}{2a} ......... {1}

which shows that equal amounts of b24ac2a \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 - 6 years, 1 month ago

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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=b2a x = \dfrac{-b}{2a} in the xyxy plane.

Venkata Karthik Bandaru - 6 years, 1 month ago

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oops i considered that to be true ........ we can write F(x) as F(x)=(x+b2a)2+b24ac4a2F(x) = (x+\frac{b}{2a} )^2 + \frac{b^2 - 4ac}{4 a^2} comparing this with the graph of f(x)=x2+kf(x) = x^2 + k which has symmetry at x=0 {the y axis} , after translation of b2a\frac{b}{2a} the symmetry shifts to x=b2ax=\frac{-b}{2a}

Abhinav Raichur - 6 years, 1 month ago

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@Abhinav Raichur Well excellent proof Abhinav Raichur !

Venkata Karthik Bandaru - 6 years, 1 month ago

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@Venkata Karthik Bandaru thank you :)

Abhinav Raichur - 6 years, 1 month ago
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