xsquare /2 -2/x square is it a polynomial?
Answers
Answer:
I have to be there answer
Answer:
Let us find the roots of the polynomial
ax2+bx+c
by completing the square. First we write the polynomial equation down, and divide both sides by the leading coefficient a:
\begin{displaymath}x^2+\frac{b}{a}x+\frac{c}{a}=0.\end{displaymath}
Then we transfer the constant term $\displaystyle \frac{c}{a} $ to the right side:
\begin{displaymath}x^2+\frac{b}{a}x=-\frac{c}{a}.\end{displaymath}
We will be using the binomial formula:
\begin{displaymath}x^2+2\cdot\frac{b}{2a}x+\left(\frac{b}{2a}\right)^2=\left(x+\frac{b}{2a}\right)^2.\end{displaymath}
Remember: always take half of the term in front of the x. Now we complete the square by adding $\displaystyle \left(\frac{b}{2a}\right)^2 $ on both sides of our equation:
\begin{displaymath}x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2.\end{displaymath}
We have completed the square! Next we use the binomial formula, and simplify the right side:
\begin{displaymath}\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.\end{displaymath}
Check the simplification on the right side! Taking square roots on both sides, we obtain
\begin{displaymath}x+\frac{b}{2a}=+\sqrt{\frac{b^2-4ac}{4a^2}}, \mbox{ or } x+\frac{b}{2a}=-\sqrt{\frac{b^2-4ac}{4a^2}}.\end{displaymath}
Solving for x, we find our two roots:
\begin{displaymath}x=-\frac{b}{2a}+\sqrt{\frac{b^2-4ac}{4a^2}}, \mbox{ or } x=-\frac{b}{2a}-\sqrt{\frac{b^2-4ac}{4a^2}}.\end{displaymath}
We obtain the classical quadratic formula by simplifying and adopting the nasty habit of using the $\pm$ notation to save us time:
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