# Polynomial Tactics

Prove that if the discriminant of a quadratic equation is less than zero, and the coefficient of $$x^2$$ is positive then $$P(x)>0$$ for all rea values of $$x$$.

In short, for polynomial $$P(x) = ax^2+bx+c$$, if $$D<0$$ and $$a>0$$, then $$P(x)>0$$ for all real values of $$x$$.

Also, if $$D<0$$ and $$a<0$$, then, $$P(x) < 0$$.

Note by Naitik Sanghavi
2 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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} $$\boxed{123}$$

Sort by:

$$P(x) = ax^{2} + bx + c$$
$$P(x) = a(x^{2} + \dfrac{b}{a}x + \dfrac{c}{a})$$
$$P(x) = a\left( \left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a^{2}}\right)$$
$$P(x) = a\left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a}$$

Now for $$a > 0$$
$$a\left(x+\dfrac{b}{2a}\right)^{2} \ge 0$$
$$a\left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a} \ge \dfrac{4ac-b^{2}}{4a}$$
$$P(x) \ge \dfrac{4ac-b^{2}}{4a}$$
$$D < 0 \rightarrow 4ac - b^{2} > 0$$
$$P(x) \ge \dfrac{4ac-b^{2}}{4a} > 0$$ for all x. Similar proof for $$a < 0$$

- 2 years, 3 months ago

Perfect!

- 2 years, 3 months ago