People who are working out the Polynomial sprint lessons may try this out also.

If \(x_1\) and \(x_2\) are non-zero roots of the equation \(ax^{2}+bx+c=0\) and \(-ax^{2}+bx+c=0\) respectively.Prove that \(\frac{ax^{2}}{2}+bx+c = 0\) has root between \(x_1\) and \(x_2\).

I put the solution in the comment box :)

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TopNewestSince \(x_1\) and \(x_2\) are roots we must have \[ax_1^{2}+bx_1+c = 0\] \[ax_2^{2}+bx_2+c\]

Now let \(f(x) = \frac{ax^{2}}{2} + bx+c\).

Hence we must have

\[f(x_1) = \frac{ax_1^{2}}{2} + bx_1+c\]---(1) \[f(x_2) = \frac{ax_2^{2}}{2} + bx_2+c\]---(2)

Adding \(\frac{ax_1^{2}}{2}\) to eqn 1 we get ,

\[f(x_1)+\frac{ax_1^{2}}{2} = ax_1^{2}+bx_1+c = 0\] \[f(x_1) = -\frac{ax_1^{2}}{2}\]

Subtracting \(\frac{3ax_1^{2}}{2}\) from eqn 2 we get , \[f(x_2)-\frac{3ax_2^{2}}{2} = -ax_2^{2}+bx_2+c = 0\] \[f(x_2) = \frac{3ax_2^{2}}{2}\]

Thus \(f(x_1\) and \(f(x_2)\) have opposite signs and hence \(f(x)\) must have a root between \(x_1\) and \(x_2\). – Eddie The Head · 2 years, 9 months ago

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Root Bounding? – Krishna Ar · 2 years, 9 months ago

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– Eddie The Head · 2 years, 9 months ago

Bounding?Indeed........Log in to reply

– Krishna Ar · 2 years, 9 months ago

:) Yes/No?Log in to reply

– Eddie The Head · 2 years, 9 months ago

YesLog in to reply