Lernmodul

In the chapter Equations - Basics you have learned basic techniques for solving linear and quadratic equations. Now let's have a look at some special types of equations, namely:

Equations with fractional expressions
Radical equations
Exponential equations

Equations with Fractions (Quotients)

In equations with fractional expressions, i.e. with quotients, the variable we want to solve for may occur also in the denominator of the fraction. Therefore, when defining the domain of the equation, we have to make sure we never divide by zero.

Example 1

$ \begin {eqnarray} \frac{6}{x-5} \,=\,2 \quad \quad \quad G\,=\,\mathbb{R}\backslash \left\{ 5 \right\} \end{eqnarray}$ (all reals except 5)

Since $\,x=5\,$ is not an element of the domain, we can multiply both sides by $\,(x-5) \neq 0\,$:

$ \begin{eqnarray} \frac{6}{x-5}\,\,&=&\,\,2 \quad \quad &\left| \,\,\cdot (x-5) \right. \\6\,&=&\,2\,(x-5) \\ 6\,&=&\,2x-10\quad \quad &\left| \,\,+10 \right. \\\,16\,\,&=&\,\,2x\quad \quad &\left| \,\,\div 2 \right.\\\,x\,\,&=&\,\,2 \end{eqnarray}$

Example 2

$ \begin{eqnarray} \frac{x-1}{2x+1}\,\,=\,\,\frac{3}{7} \end{eqnarray}$

For x = –0.5 the denominator equals 0. Therefore the value -0.5 has to be excluded from the domain:

$ G\,\,=\,\,\mathbb{R}\backslash \left\{ -0.5 \right\} $

The equation has to be multiplied by the least common multiple (LCM) of the two denominator terms:

$ \begin{eqnarray} \frac{x-1}{2x+1}\,\,&=&\,\,\frac{3}{7} \quad \quad &\left| \,\,\cdot 7 \cdot (2x+1) \right. \\7x-7\,&=&\,6x+3 \quad \quad &\left| \,\,-6x+7 \right. \\ x&=&10 \end{eqnarray}$

Example 3

$\begin{eqnarray}\frac{3x+3}{2x-16}-4+\frac{2x+2}{x-8}=\frac{3x-3}{x-8} \end{eqnarray} \quad \quad \quad G=\mathbb{R}\backslash \left\{ 8 \right\}$

For x = 8 the denominator equals 0. The LCM of the denominators is 2x – 16 = 2(x – 8).

$\begin{eqnarray} \frac{3x+3}{2x-16}-4+\frac{2x+2}{x-8}&=& \frac{3x-3}{x-8} \quad &\left| \;\cdot \,(2x-16) \right. \\ 3x+3-4\cdot (2x-16)+2\cdot (2x+2)&=&2\cdot (3x-3) \quad \\ 3x+3-8x+64+4x+4&=&6x-6 \\ -x+71&=&6x-6 \quad &\left| \; -6x-71 \right. \\ -7x&=&-77 \quad &\left| \,\div (-7) \right. \\ x&=&11 \end{eqnarray}$

Radical Equations

Equations containing radical expressions, i.e. roots, are called radical equations.

We focus here on equations with square roots. These are typically solved by squaring both sides of the equation in order to get rid of the radical expression. But you have to be aware of two important facts:

Squaring both sides of an equation may enlarge the solution set of an equation!
It is possible that some of the solutions we find at the end of the squaring process are not valid solutions of the initial equation (we call them extraneous solutions). Therefore you must always check the "solutions" after having solved a radical equation, by inserting them into the initial equation!
A radical can be eliminated successfully only if it is isolated before getting squared. If the radical is part of a sum, squaring doesn't eliminate it! (Remember that you have to square sides, not terms. This means that the squaring process runs according to the Binomial Formulas.)

Example 4

$ \begin{eqnarray} \sqrt{x+2} +4\,&=&\,x \end{eqnarray}$

The term under the radical sign must not be negative. Therefore the domain of the equation ist limited:

$ G\,=\,\left \{ x \in \mathbb{R} \,\left | \,x \geq -2 \right. \right\} $

To get rid of the radical, we have to isolate it:

$ \begin{eqnarray} \sqrt{x+2} +4\,&=&\,x &\left|\;-4\right. \\ \sqrt{x+2}\,&=&\,x-4 \quad &\left|\; \text{square both sides}\right.\\ x+2 &=& {(x-4)^2}\end{eqnarray}$

Now the right hand side can be expanded according to the Binomial Formulas:

$ \begin{eqnarray} x+2&=&{x^2}-8x+16 \quad &\left|\;-x-2 \right.\\ 0&=&{x^2}-9x+14 \end{eqnarray}$

This quadratic equation can be solved by applying the Quadratic Formula (see also the paragraph "Quadratic Equations" in the chapter Equations - Basics):

${x_{1,2}} ={\large \frac{ 9 \pm \sqrt {81-56}}{2}} = {\large \frac{9 \pm 5}{2}} \quad \qquad {x_1}=7 \;\quad {x_2}=2 $

If you insert both values into the initial equation you will notice that $\,{x_1}=7\,$ is a valid solution, but $\,{x_2}=2\,$ is not:

$ \begin{eqnarray} \sqrt{7+2} +4\,&=&\,7 \quad \end{eqnarray}$ is true

$ \begin{eqnarray} \sqrt{2+2} +4\,&=&\,2 \quad \end{eqnarray}$ is false!

What happened? The last equation before squaring both sides was:

$ \sqrt{x+2}\,=\,x-4$

If you substitute 2 for x in the equation, you get a wrong statement:

$\sqrt{2+2}\,\neq \,2-4$

But: the two sides differ only by sign. By squaring, the minus sign vanishes, and you suddenly have a "solution" of the new equation! Obviously the squaring process has changed the solution set of the equation! The final equation isn't equivalent to the initial one.

Therefore the solution set of the original equation is: $L\,=\,\left\{ 7 \right\}$

Example 5

$ \begin{eqnarray} \sqrt{4x+9} -2\,&=&\,\sqrt{3x-5} \end{eqnarray}$

The expressions under the square root sign must not be negative. Thereby, the condition $\;3x-5 \geq 0\;$ is more restrictive than $\;4x+9 \geq 0\;$; hence the domain of the equation is

$ G\,=\,\left \{ x \in \mathbb{R} \,\left | \,x \geq \frac {5}{3} \right. \right\} $

In this example we cannot isolate both radicals simultaneously. We have to eliminate the radicals one at a time. Remember, too, that you have to square sums by use of the binomial formulas:

$ \begin{eqnarray} \sqrt{4x+9} -2\,&=&\,\sqrt{3x-5} &\left|\;\text{square both sides}\right. \\ 4x+9 - 2 \cdot 2 \cdot \sqrt{4x+9} + 4\,&=&\,3x-5\quad &\left|\;-3x+5+4\,\sqrt{4x+9} \right.\\ x+18 &=& 4\,\sqrt{4x+9}&\left|\;\text{square both sides}\right.\\ {x^2}+36x+324 &=& 64x+144&\left|\;-64x-144\right.\\{x^2}-28x+180 &=& 0\end{eqnarray}$

The solutions of this quadratic equation are:

${x_{1,2}} ={\large \frac{ 28 \pm \sqrt {784-720}}{2}} = {\large \frac{28 \pm 8}{2}} \quad \qquad {x_1}=18 \;\quad {x_2}=10 $

Check each solution in the original equation:

$ \begin{eqnarray} \sqrt{40+9} -2\,&=&\,\sqrt{30-5} \quad \end{eqnarray}$ is true: $\; 7=7$

$ \begin{eqnarray} \sqrt{72+9} -2\,&=&\,\sqrt{54-5} \quad \end{eqnarray}$ is true: $\; 5=5$

Both values are valid solutions of the initial equation.

$L\,=\,\left\{10;\,18 \right\}$

Exponential Equations

An equation where the (unknown) variable appears in the exponent is called an exponential equation.

Examples

${2^x}=3$

$100 \cdot {1.04^x}=200$

When solving such equations we often use a third category of manipulations. It again has the equivalence preserving property, as described in the chapter Equations - Basics.

E3: Take the logarithm of both sides of an equation.

Example 6

\[\begin{align} & {{10}^{x}}\,\,=\,\,1000 \\ & \log {{10}^{x}}\,\,=\,\,\log 1000 \\ & x\cdot \log 10\,\,=\,3 \\ & x\,\,=\,\,3 \end{align}\]

In the second step we applied a logarithmic law (see chapter Logarithm):

$\log {{a}^{k}}\,\,=\,\,k\cdot \log a$

You can solve the example in an easier way:

$\begin{align} & {{10}^{x}}\,\,=\,\,{{10}^{3}} \\ & x\,\,=\,\,3 \end{align}$

By converting the 1000 to a power of 10 on the right-hand side of the equation, we get equal bases, so the same must hold for the exponents!

In the following example we make use of the logarithmic law again :

Example 7