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).
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2 \times 3
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2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
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consider the function to be F(x)=ax2+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=2a−b+or−2ab2−4ac ......... {1}
which shows that equal amounts of 2ab2−4ac 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=2a−b in the xy plane.
oops i considered that to be true ........ we can write F(x) as
F(x)=(x+2ab)2+4a2b2−4ac comparing this with the graph of f(x)=x2+k which has symmetry at x=0 {the y axis} , after translation of 2ab the symmetry shifts to x=2a−b
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
consider the function to be F(x)=ax2+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=2a−b+or−2ab2−4ac ......... {1}
which shows that equal amounts of 2ab2−4ac 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=2a−b in the xy plane.
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oops i considered that to be true ........ we can write F(x) as F(x)=(x+2ab)2+4a2b2−4ac comparing this with the graph of f(x)=x2+k which has symmetry at x=0 {the y axis} , after translation of 2ab the symmetry shifts to x=2a−b
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