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# Polynomial Sprint Extra

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 :)

Note by Eddie The Head
3 years, 8 months ago

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## Comments

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Since $$x_1$$ 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 $$x_2$$.

- 3 years, 8 months ago

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Root Bounding?

- 3 years, 8 months ago

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Bounding?Indeed........

- 3 years, 8 months ago

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:) Yes/No?

- 3 years, 8 months ago

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Yes

- 3 years, 8 months ago

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