**Note**For this tutorial, it is assumed that you know how to solve linear and quadratic equations. If you feel you need to review this, go back to %4. Equations often arise in calculus that are not polynomial equations of low degree. Many of these complicated-looking equations can be solved easily using a few basic techniques to first write them in the form $P \cdot Q = 0$ or $P/Q = 0,$ where $P$ and $Q$ are simpler expressions.

Equations that reduce to the form

If we can manipulate a complicated-looking equation into the form $P \cdot Q = 0$ then we can solve it easily we you remember the following, which we used in the preceeding topic:
*P*·*Q*= 0**Solving equations that reduce to the form**

*P*·*Q*= 0We use the fact that, if a product is equal to 0, then at least one of the factors must be 0. That is, if

- $P \cdot Q = 0$,

- $P = 0$ or $Q = 0.$

**Note**This argument applies as well to a product of three or more terms; for instance, if

- $P \cdot Q \cdot R = 0$,

- $P = 0$, $Q = 0$, or $R = 0$.

**Examples**

**1.**\t $4x^7-x^5 = 0$ \gap[10] \\ \t $x^5(4x^2-1)=0$ \gap[10] \t Factor the left-hand side. \\ \t Either $x^5=0$ or $(4x^2-1)=0.$ \gap[10] \t Either $P=0$ or $Q=0.$ \\ \t $x=0, x=-\frac{1}{2}$ %or $x=\frac{1}{2}$ \gap[10] \t Solve the individual equations. \\ \\

^{ }**2.**\t $(2x+1)(x^2-4)-(x-3)(x^2-4) = 0$ \gap[10] \\ \t $[(2x+1) - (x-3)](x^2-4) = 0$ \gap[10] \t Common factor $(x^2-4)$ \\ \t $(x+4)(x^2-4) = 0$ \gap[10] \t Simplify. \\ \t $(x+4)(x-2)(x+2) = 0$ \gap[10] \t Factor the quadratic. \\ \t $x=-4, x=2$ %or $x=-2$ \gap[10] \t $P = 0$, $Q = 0$, %or $R = 0$ \\ \\

^{ }**3.**\t $x\sqrt{2x-1} = \sqrt{2x-1}$ \gap[10] \\ \t $x\sqrt{2x-1} - \sqrt{2x-1} = 0$ \\ \t $\sqrt{2x-1}(x - 1) = 0$ \gap[10] \t Common factor $\sqrt{2x-1}$ \\ \t Either $\sqrt{2x-1}=0$ or $x-1=0.$ \gap[10] \t Either $P=0$ or $Q=0.$ \\ \t Either $2x-1=0$ or $x-1=0.$ \gap[10] \t %If $\sqrt{a} = 0,$ %then $a = 0$ \\ \t $x=\frac{1}{2}$ %or $x=1$ \gap[10] \t Solve the individual equations.

^{ }**Some for you to do**Solve the following equations. Solutions must be exact (not decimal approximations). Enter each solution as a comma-separated list, for instance -2, -1/2, 3.

Equations that reduce to the form

If we can manipulate a complicated-looking equation into the form $P / Q = 0$ then we can solve it using the following observation:
*P*/*Q*= 0**Solving equations that reduce to the form**

*P*/*Q*= 0We use the fact that, if $\displaystyle \frac{P}{Q}=0$, then $P=0$.

**Note**

The expression $\dfrac{P}{Q}$ is not defined if $Q = 0$ so you need to eliminate all solutions that make $Q$ zero.

**Examples**

**1.**\t $\displaystyle \frac{x^2-1}{x-2} = 0$ \gap[10] \\ \t $x^2-1=0$ \gap[10] \t If $\dfrac{P}{Q}=0$ then $P=0$. \\ \t $(x-1)(x+1) = 0$ \gap[10] \t Factor. \\ \t $x=-1$ %or $x=1$ \gap[10] \t Neither solution makes the denominator of the original expression zero, so we accept both. \\ \\

**2.**\t $\displaystyle \frac{(x+1)(x+2)^2 - (x+1)^2(x+2)}{(x+1)^4 - (x-2)^2} = 0$ \gap[10] \\ \t $(x+1)(x+2)^2 - (x+1)^2(x+2) = 0$ \gap[10] \t If $\dfrac{P}{Q}=0$ then $P=0$. \\ \t $(x+1)(x+2)[(x+2) - (x+1)] = 0$ \gap[10] \t Factor. \\ \t $(x+1)(x+2)(1) = 0$ \\ \t $x = -1$ %or $x = -2$ \t Neither solution makes the denominator of the original expression zero, so we accept both. \\ \\

**3.**\t $\displaystyle \frac{(x+1)(x+2)^2 - (x+1)^2(x+2)}{x+2} = 0$ \gap[10] \\ \t $(x+1)(x+2)^2 - (x+1)^2(x+2) = 0$ \gap[10] \t If $\dfrac{P}{Q}=0$ then $P=0$. \\ \t $x = -1$ %or $x = -2$ \gap[10] \t We solved this equation in the preceding example. \\ \t $x = -1$ \t We reject $x = -2$ as it makes the denominator of the original expression zero, so $x$ cannot be $-2$ in the original equatino.

**Some for you to do**Solve the following equations. Solutions must be exact (not decimal approximations). Enter each solution as a comma-separated list, for instance -2, -1/2, 3.

Combining techniques

In the following you will need to use the above techniques as well as various things you learned about rational functions in the %3.
Now try the exercises in Section 0.6 of or , or move ahead to the next tutorial by pressing on the sidebar.

*September, 2017*

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