Modulus Equations with Restrictions | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

The right side must be non-negative

In $|u|=v$ , an expression such as $2x+1$ is not automatically positive. First require $v\ge0$ ; then solve $u=v$ or $u=-v$ .

Worked example

Solve |x − 4| = 2x + 1.

Require $2x+1\ge0$ , so $x\ge-1/2$ . The two equations produce candidates: possible roots that still need checking.

\begin{gathered}x-4=2x+1\\x=-5\\[4pt]x-4=-(2x+1)\\3x=3,\quad x=1\end{gathered}

Reject −5: it violates the restriction and gives $9=-9$ in the original equation. At 1, both sides equal 3. The only solution is $x=1$ .

Common mistake

The minus sign acts on the whole right side:

-(2x+1)=-2x-1

, not

-2x+1

Your turnPractice

Solve $|x+2|=2x-1$ . Show why one candidate must be rejected.

Show worked answer

Require $x\ge1/2$ .

\begin{gathered}x+2=2x-1\\x=3\\[4pt]x+2=-2x+1\\x=-1/3\end{gathered}

Answer

x=3

Reject $-1/3$ because the right side is negative. At 3, both sides equal 5.

Keep a denominator away from zero

A fraction inside the bars uses the same two-case rule. Record any forbidden input before multiplying by its denominator.

Worked example

Solve |(x + 1)/(x − 2)| = 2.

$x\ne2$ , since division by zero is undefined.

\begin{gathered}\frac{x+1}{x-2}=2\\x+1=2x-4\\x=5\\[6pt]\frac{x+1}{x-2}=-2\\x+1=-2x+4\\x=1\end{gathered}

Neither root is 2; the original fractions are 2 and −2.

Your turnPractice

Solve $\left|\dfrac{2x-1}{x+1}\right|=1$ , stating the excluded input.

Show worked answer

Exclude $x=-1$ . Solve $2x-1=x+1$ or $2x-1=-(x+1)$ .

Answer

x=2\quad\text{or}\quad x=0

Both satisfy the original equation and neither is excluded.

The right side must be non-negative

In $|u|=v$ , an expression such as $2x+1$ is not automatically positive. First require $v\ge0$ ; then solve $u=v$ or $u=-v$ .

Worked example

Solve |x − 4| = 2x + 1.

Require $2x+1\ge0$ , so $x\ge-1/2$ . The two equations produce candidates: possible roots that still need checking.

\begin{gathered}x-4=2x+1\\x=-5\\[4pt]x-4=-(2x+1)\\3x=3,\quad x=1\end{gathered}

Reject −5: it violates the restriction and gives $9=-9$ in the original equation. At 1, both sides equal 3. The only solution is $x=1$ .

Common mistake

The minus sign acts on the whole right side:

-(2x+1)=-2x-1

, not

-2x+1

Your turnPractice

Solve $|x+2|=2x-1$ . Show why one candidate must be rejected.

Show worked answer

Require $x\ge1/2$ .

\begin{gathered}x+2=2x-1\\x=3\\[4pt]x+2=-2x+1\\x=-1/3\end{gathered}

Answer

x=3

Reject $-1/3$ because the right side is negative. At 3, both sides equal 5.

Keep a denominator away from zero

A fraction inside the bars uses the same two-case rule. Record any forbidden input before multiplying by its denominator.

Worked example

Solve |(x + 1)/(x − 2)| = 2.

$x\ne2$ , since division by zero is undefined.

\begin{gathered}\frac{x+1}{x-2}=2\\x+1=2x-4\\x=5\\[6pt]\frac{x+1}{x-2}=-2\\x+1=-2x+4\\x=1\end{gathered}

Neither root is 2; the original fractions are 2 and −2.

Your turnPractice

Solve $\left|\dfrac{2x-1}{x+1}\right|=1$ , stating the excluded input.

Show worked answer

Exclude $x=-1$ . Solve $2x-1=x+1$ or $2x-1=-(x+1)$ .

Answer

x=2\quad\text{or}\quad x=0

Both satisfy the original equation and neither is excluded.