Roots of Equations

Equations of the Form \(\sqrt{ax^{2}+bx+c}\pm\sqrt{ax^{2}+bx+d}=k\)

\begin{equation*} \sqrt{ax^{2}+bx+c}+\sqrt{ax^{2}+bx+d}=k \end{equation*}

Let \(A=\sqrt{ax^{2}+bx+c}\) and \(B=\sqrt{ax^{2}+bx+d}\). Then \(A+B=k\).

But note that \(A^{2}-B^{2}\) is a constant (\(c-d\))! So we know:

\begin{equation*} \frac{A^{2}-B^{2}}{A+B}=\frac{c-d}{k}=A-B \end{equation*}

We now know \(A+B\) and \(A-B\). Solve for one of them and get \(x\) from there.

\((x+a)(x+b)(x+c)(x+d)=k\) where \(a+b=c+d=m\)

\begin{equation*} (x^{2}+(a+b)x+ab)(x^{2}+(c+d)x+cd)=k \end{equation*}

\begin{equation*} (x^{2}+mx+ab)(x^{2}+mx+cd)=k \end{equation*}

Let \(y=x^{2}+mx\) and solve for \(y\).

\(ax^{4}+bx^{3}+cx^{2}+bx+a=0\)

Factor:

\begin{equation*} a\left(x^{4}+1\right)+bx\left(x^{2}+1\right)+cx^{2}=0 \end{equation*}

Divide by \(x^{2}\):

\begin{equation*} a\left(x^{2}+\frac{1}{x^{2}}\right)+b\left(x+\frac{1}{x}\right)+c=0 \end{equation*}

Let \(y=x+\frac{1}{x}\) and note that

\begin{equation*} y^{2}-2=x^{2}+\frac{1}{x^{2}} \end{equation*}

Solve for \(y\).

Showing a Polynomial Has no Real Roots

If we can show that \((x-c)\) is not a factor of a polynomial for any \(c\), then it has no real roots. This is sometimes obvious just upon observation: We know from the Remainder Theorem that when divided by \((x-c)\), the remainder is \(p(c)\). So for a root to exist, \(p(c)\) must be 0. But if all the coefficients are positive and of even power, then it’s easy to see that it will not be 0 for any value of \(x\).