Polynomial Equations with Changed Inputs | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

First solve for the whole input

To solve $P(g(x))=0$ , first find every root of $P(t)=0$ . For each root $t=r$ , solve $g(x)=r$ . Here $t$ is just a name for the input to $P$ . Check the range of $g$ : some original roots may be impossible for the new input.

Worked example

P(t) = (t + 1)(t − 2)(t − 3). Solve P(|x|) = 0.

Set the input $|x|$ equal to each root of $P$ : −1, 2 or 3. Modulus cannot be negative, so discard −1.

|x|=2\ \text{or}\ |x|=3

Answer

x=\pm2,\qquad x=\pm3

For $P(x^2)=0$ instead, require $x^2=2$ or $x^2=3$ , giving $x=\pm\sqrt2,\ \pm\sqrt3$ .

Your turnPractice

$P(t)=t(t-4)(t+2)$ . Solve $P(|x-1|)=0$ and $P(x^2)=0$ . Count distinct solutions.

Show worked answer

The input can be 0, 4 or −2, but both new inputs are non-negative. Keep only 0 and 4.

|x-1|=0\ \text{or}\ |x-1|=4

Answer

x=1,\ -3,\ 5

For squared input, $x^2=0$ or $x^2=4$ gives $x=0,\ -2,\ 2$ . Each equation has three distinct solutions; zero is not two different square roots.

Name a repeated expression before solving

Worked example

Solve |x² − 5| = 4.

The whole inside is either 4 or −4. Each case is now a quadratic, so two sign cases need not mean only two roots.

\begin{aligned}x^2-5=4&\Rightarrow x=\pm3,\\x^2-5=-4&\Rightarrow x=\pm1.\end{aligned}

All four roots give modulus 4. There is no need to sketch a nonlinear modulus graph.

Worked example

Solve |x − 1|² + |x − 1| − 6 = 0.

Let $u=|x-1|$ , with $u\ge0$ . This turns the equation into a quadratic:

\begin{gathered}u^2+u-6=0\\(u+3)(u-2)=0\end{gathered}

Reject $u=-3$ , then solve $|x-1|=2$ .

Answer

x=-1\quad\text{or}\quad x=3

Substitute either root into the original expression: $4+2-6=0$ .

Your turnPractice

Solve $|x+2|^2-3|x+2|-4=0$ . State the restriction on your substitution.

Show worked answer

Let $u=|x+2|\ge0$ . Then $(u-4)(u+1)=0$ ; keep 4 and reject −1.

Answer

x=2\quad\text{or}\quad x=-6

First solve for the whole input

Worked example

P(t) = (t + 1)(t − 2)(t − 3). Solve P(|x|) = 0.

Set the input $|x|$ equal to each root of $P$ : −1, 2 or 3. Modulus cannot be negative, so discard −1.

|x|=2\ \text{or}\ |x|=3

Answer

x=\pm2,\qquad x=\pm3

For $P(x^2)=0$ instead, require $x^2=2$ or $x^2=3$ , giving $x=\pm\sqrt2,\ \pm\sqrt3$ .

Your turnPractice

$P(t)=t(t-4)(t+2)$ . Solve $P(|x-1|)=0$ and $P(x^2)=0$ . Count distinct solutions.

Show worked answer

The input can be 0, 4 or −2, but both new inputs are non-negative. Keep only 0 and 4.

|x-1|=0\ \text{or}\ |x-1|=4

Answer

x=1,\ -3,\ 5

For squared input, $x^2=0$ or $x^2=4$ gives $x=0,\ -2,\ 2$ . Each equation has three distinct solutions; zero is not two different square roots.

Name a repeated expression before solving

Worked example

Solve |x² − 5| = 4.

The whole inside is either 4 or −4. Each case is now a quadratic, so two sign cases need not mean only two roots.

\begin{aligned}x^2-5=4&\Rightarrow x=\pm3,\\x^2-5=-4&\Rightarrow x=\pm1.\end{aligned}

All four roots give modulus 4. There is no need to sketch a nonlinear modulus graph.

Worked example

Solve |x − 1|² + |x − 1| − 6 = 0.

Let $u=|x-1|$ , with $u\ge0$ . This turns the equation into a quadratic:

\begin{gathered}u^2+u-6=0\\(u+3)(u-2)=0\end{gathered}

Reject $u=-3$ , then solve $|x-1|=2$ .

Answer

x=-1\quad\text{or}\quad x=3

Substitute either root into the original expression: $4+2-6=0$ .

Your turnPractice

Solve $|x+2|^2-3|x+2|-4=0$ . State the restriction on your substitution.

Show worked answer

Let $u=|x+2|\ge0$ . Then $(u-4)(u+1)=0$ ; keep 4 and reject −1.

Answer

x=2\quad\text{or}\quad x=-6