Waste less time on Facebook — follow Brilliant.
×

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, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 - 3 years, 5 months ago

Log in to reply

Root Bounding?

Krishna Ar - 3 years, 5 months ago

Log in to reply

Bounding?Indeed........

Eddie The Head - 3 years, 5 months ago

Log in to reply

:) Yes/No?

Krishna Ar - 3 years, 5 months ago

Log in to reply

@Krishna Ar Yes

Eddie The Head - 3 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...