Transformation of Equations

Sometimes the most efficient way to solve or analyse an equation is not to attack it directly, but to transform it into a related equation whose roots are a known function of the original roots (shifted, scaled, sign-flipped, or inverted), solve or study that instead, and translate the answer back. Four transformations come up constantly:

• Shift (translate) the roots by $k$: if $\alpha_1,\dots,\alpha_n$ are the roots of $f(x)=0$, then $\alpha_1+k,\dots,\alpha_n+k$ are exactly the roots of $f(x-k)=0$. (Substituting $x \to x-k$ pushes every root up by $k$.) This is the standard trick for killing the second-highest term of an equation — choosing $k$ appropriately makes the new $x^{n-1}$ coefficient vanish.
• Flip the sign of every root: the roots of $f(-x)=0$ are exactly $-\alpha_1,\dots,-\alpha_n$. Concretely, $f(-x)$ is obtained from $f(x)$ by flipping the sign of every other coefficient (starting from the second).
• Scale every root by $k$: the roots of $k^n f(x/k) = 0$ are exactly $k\alpha_1,\dots,k\alpha_n$ (choose $k$ to clear any fractions this substitution introduces).
• Invert every root: the roots of $x^n f(1/x) = 0$ are exactly $1/\alpha_1,\dots,1/\alpha_n$ (valid whenever no root is zero). This transformation — reversing the coefficient list — is the key idea behind reciprocal equations, taken up next.

Each of these can be carried out either by direct algebraic substitution, or, when only a shift is needed, by the same synthetic-division bookkeeping from Section 4.4, applied repeatedly.

Worked example. Find the equation whose roots are $2$ more than the roots of $x^3 - 9x^2 + 23x - 15 = 0$ (whose roots happen to be $1,3,5$).

By the shift rule with $k=2$, the required equation is $f(x-2)=0$, i.e. substitute $x \to x-2$:

$(x-2)^3 - 9(x-2)^2 + 23(x-2) - 15 = 0.$ …