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 :)

#Algebra #Polynomials #CosinesGroup #TorqueGroup #PolynomialSprint

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Comments

Root Bounding?

Krishna Ar - 6 years, 11 months ago

Bounding?Indeed........

Eddie The Head - 6 years, 11 months ago

:) Yes/No?

Krishna Ar - 6 years, 11 months ago

@Krishna Ar – Yes

Eddie The Head - 6 years, 11 months ago

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 $x_2$ .

Eddie The Head - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$