Other types of equations

Example 1: Solve

${\dfrac{1}{x}+\dfrac{1}{x+2}=\dfrac{3}{4}}$

Solution:

The denominators are ${0}$ when ${x=0}$ or ${x=-2}$. These cannot be solutions.

The LCD is ${4x(x+2)}$. Multiply every term by it. Each denominator cancels:

$\begin{align*}4(x+2)+4x&=3x(x+2)\\8x+8&=3x^2+6x\end{align*}$

This is quadratic. Move every term to the right side, and factor:

$\begin{align*}0&=3x^2-2x-8\\0&=(3x+4)(x-2)\end{align*}$

So ${x=-\dfrac{4}{3}}$ or ${x=2}$. Neither is ${0}$ or ${-2}$, so both are solutions.

Check ${x=2}$:

${\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}}$

It checks.

Example 2: Solve ${\dfrac{x}{x-3}=\dfrac{3}{x-3}+2}$.

Solution:

The denominator is ${0}$ when ${x=3}$, so ${3}$ cannot be a solution. Multiply every term by ${x-3}$:

$\begin{align*}x&=3+2(x-3)\\x&=2x-3\\x&=3\end{align*}$

But ${x=3}$ makes the denominators ${0}$. It is an extraneous solution. So the equation has no solution.

Example 3: One pipe can fill a tank in ${6}$ hours. A second, larger pipe can fill it in ${3}$ hours. How long does it take to fill the tank with both pipes open?

Solution:

Think about how much of the tank each pipe fills in one hour. The first pipe fills the tank in ${6}$ hours, so it fills ${\dfrac{1}{6}}$ of the tank each hour. The second pipe fills ${\dfrac{1}{3}}$ of the tank each hour.

Let ${t}$ be the time, in hours, with both pipes open. Together they fill ${\dfrac{1}{t}}$ of the tank each hour. The amounts per hour add up:

${\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{t}}$

Multiply every term by the LCD, ${6t}$:

$\begin{align*}t+2t&=6\\3t&=6\\t&=2\end{align*}$

With both pipes open, the tank fills in ${2}$ hours. This makes sense: it is faster than either pipe alone.

Example 4: Solve each equation.

(a) ${\sqrt{2x+3}=x}$

(b) ${\sqrt{x+5}-\sqrt{x}=1}$

Solution:

(a) The radical is already alone. Square both sides:

$\begin{align*}2x+3&=x^2\\0&=x^2-2x-3\\0&=(x-3)(x+1)\end{align*}$

So ${x=3}$ or ${x=-1}$. Check each one in the original equation:

• ${x=3}$: ${\sqrt{9}=3}$. True.
• ${x=-1}$: ${\sqrt{1}=1}$, but the right side is ${-1}$. False.

So ${-1}$ is extraneous. The only solution is ${x=3}$.

(b) Get one radical alone first:

${\sqrt{x+5}=1+\sqrt{x}}$

Square both sides. On the right, use ${(a+b)^2=a^2+2ab+b^2}$:

${x+5=1+2\sqrt{x}+x}$

A radical is still left. Get it alone, then square again:

$\begin{align*}4&=2\sqrt{x}\\2&=\sqrt{x}\\4&=x\end{align*}$

Check: ${\sqrt{9}-\sqrt{4}=3-2=1}$. So ${x=4}$.

Example 5: Solve ${x^{2/3}=4}$.

Solution:

Cube both sides. Since ${\left(x^{2/3}\right)^3=x^{2}}$:

${x^2=64}$

So ${x=8}$ or ${x=-8}$.

Check: ${8^{2/3}=\left(\sqrt[3]{8}\right)^2=2^2=4}$, and $(-8)^{2/3}=\left(\sqrt[3]{-8}\right)^2=(-2)^2=4$. Both are solutions.

Example 6: Solve each equation.

(a) ${|2x-5|=7}$

(b) ${|x+1|+4=2}$

Solution:

(a) The expression ${2x-5}$ is ${7}$ or ${-7}$. Solve both equations:

${2x-5=7}$   or   ${2x-5=-7}$

${2x=12}$   or   ${2x=-2}$

So ${x=6}$ or ${x=-1}$.

(b) Get the absolute value alone first:

${|x+1|=-2}$

An absolute value is never negative. So there is no solution.

Example 7: Solve each equation.

(a) ${x^4-5x^2+4=0}$

(b) ${x-5\sqrt{x}+6=0}$

Solution:

(a) Let ${u=x^2}$. Then ${x^4=u^2}$, and the equation becomes

$\begin{align*}u^2-5u+4&=0\\(u-1)(u-4)&=0\end{align*}$

So ${u=1}$ or ${u=4}$. Go back to ${x}$: ${x^2=1}$ or ${x^2=4}$.

So there are four solutions: ${x=\pm 1}$ and ${x=\pm 2}$.

(b) Let ${u=\sqrt{x}}$. Then ${x=u^2}$, and the equation becomes

$\begin{align*}u^2-5u+6&=0\\(u-2)(u-3)&=0\end{align*}$

So ${\sqrt{x}=2}$ or ${\sqrt{x}=3}$. That is, ${x=4}$ or ${x=9}$.

Check ${x=9}$: ${9-5(3)+6=0}$. Also, ${x=4}$: ${4-5(2)+6=0}$. Both check.

Example 8: Solve each equation.

(a) ${x^3=4x}$

(b) ${x^3-2x^2-9x+18=0}$

Solution:

(a) Do not divide by ${x}$: that would lose a solution. Move every term to one side, and factor:

$\begin{align*}x^3-4x&=0\\x(x^2-4)&=0\\x(x+2)(x-2)&=0\end{align*}$

So ${x=0}$, ${x=-2}$, or ${x=2}$.

(b) Four terms, so factor by grouping:

$\begin{align*}x^2(x-2)-9(x-2)&=0\\(x-2)(x^2-9)&=0\\(x-2)(x+3)(x-3)&=0\end{align*}$

So ${x=2}$, ${x=-3}$, or ${x=3}$.