Solving some non-linear equations

There are other types of equations, non-linear ones, which we can solve as well. There are many more non-linear equations, some of which we will learn to solve much later.

Equations with a square term

These are equations of the form

\boxed{a x^2+b=0}

For example,

3x^2-12=0

Note that this equation is similar to the linear equation, but the variable $x$ is squared. To solve this kind of equation:

bring all the $x^2$ -terms on one side, and the number on the other side
isolate the $x^2$
take the root on both sides of the equation sign

\begin{array}{llll} 3x^2 -12&= &0 &\quad| +12\\ 3x^2 &= &12 &\quad| :3\, (\text{isolate})\\ x^2 &= &4 &\quad|\sqrt{\phantom{x}}\\ \sqrt{x^2} &= &\sqrt{4} &\quad| \text{simplify} \\ x &= &\pm \sqrt{4} = \pm 2 \end{array}

Note that $-2$ is also a solution, because we also have $(-2)^2=4$ . So we have two solutions, $x=2$ and $x=-2$ .

Sometimes with have to work a bit harder to isolate the $x^2$ . For example, consider following example:

Example 1

Solve the equation

3x^2-4= x^2+5

Again, we solve it by first bringing all $x^2$ terms on one side, and all the numbers on the other side.

\begin{array}{llll} 3x^2-4&= &x^2+5 &\quad| -x^2, +4\\ 3x^2-x^2 &= &5+4 &\quad| \text{simplify}\\ 2x^2 &= &9 &\quad| :2\, (\text{isolate})\\ x^2 &= & 4.5 &\quad| \sqrt{\phantom{x}}\\ \sqrt{x^2} &= & \sqrt{4.5} &\quad|\text{simplify}\\ x &= &\pm \sqrt{4.5} = \pm2.121 \end{array}

Note that we write $x=\pm \sqrt{4.5}$ in order to get both solutions.

Sometimes, an equation does not have a solution. Have a look at the following example:

Example 2

Solve the equation

x^2+4=0

We try to solve it as usual:

\begin{array}{llll} x^2+4&= &0 &\quad| -4\\ x^2 &= & -4 &\quad| \sqrt{\phantom{x}}\\ \sqrt{x^2} &= &\sqrt{-4} &\\ \end{array}

and because $\sqrt{-4}$ does not exist, we must conclude that there is no value for which this equation holds.

Equations with a square root term

These are equations of the form

\boxed{a\sqrt{x}+b=0}

For example,

3\sqrt{x}-10=8

We use a similar approach as above:

bring all the $\sqrt{x}$ -terms on one side, and the number on the other side
isolate the $\sqrt{x}$
square both sides

Thus, we have

\begin{array}{llll} 3\sqrt{x}-10&= &8 &\quad| +10\\ 3\sqrt{x} &= &18 &\quad| :3\, (\text{isolate})\\ \sqrt{x} &= & 6 &\quad| \text{square both sides}\\ \left(\sqrt{x}\right)^2 &= & 6^2 & \quad|\text{simplify}\\ x &= & 36 &\\ \end{array}

However, sometimes you find a value for $x$ , but it is not a solution of the equation. Have a look at the following example:

Example 3

Solve the equation

3\sqrt{x}+10 = 1

Solving as usual, we get

\begin{array}{llll} 3\sqrt{x}+10&= &1 &\quad| -10\\ 3\sqrt{x} &= &-9 &\quad| :3\, (\text{isolate})\\ \sqrt{x} &= &-3 &\quad| \text{square both sides}\\ \left(\sqrt{x}\right)^2 &= & (-3)^2 & \quad|\text{simplify}\\ x &= & 9 &\\ \end{array}

In fact, this equation has no solution - so why do we get one? Well, look at the third line, where it says $\sqrt{x}=-3$ . No such $x$ -value exists, because the root is always positive.

Equations with x in the denominator

These are equations of the following type:

\boxed{a\frac{1}{x}+b=0}

or, written a bit more compact:

\boxed{\frac{a}{x}+b=0}

An example is the following equation:

\frac{4}{x}-2=0

To solve it, follow the steps:

bring all $\frac{1}{x}$ -terms on one side, and all numbers on the other side of the equal sign
Multiply both sides by $x$
Solve the resulting equation

So let's do this:

\begin{array}{llll} \frac{4}{x}-2&= &0 &\quad| +2\\ \frac{4}{x} &= &2 &\quad| \cdot x\\ 4 &= & 2x &\quad| :2\\ 2 &= & x &\\ \end{array}

Sometimes you have to work for bringing the equation in this form, and sometimes the denominator is not just $x$ . But we apply the same strategy. Here is an example.

Example 4

Solve the equation

\frac{3}{x}-1 = \frac{2}{3x}+\frac{1}{3}

Again, let's bring all fractions with $x$ in the denominator on one side, and write them as one fraction. Same for the numbers. Then multiply by $x$ and solve the equation.

\begin{array}{llll} \frac{3}{x}-1&=& \frac{2}{3x}+\frac{1}{3} &\quad| +1, -\frac{2}{3x}\\ \frac{3}{x}-\frac{2}{3x} &= &\frac{1}{3}+1 &\quad| \text{write as a fraction}\\ \frac{7}{3x} &= & \frac{4}{3} &\quad| \cdot x\\ \frac{7}{3} &= & \frac{4}{3}x &\quad| \cdot \frac{3}{4}\\ \frac{21}{12} &= & x &\\ \end{array}

The result is $x=\frac{21}{12}=\frac{7}{4}$ . Of course we could have multiplied both sides by $3x$ , rather than just by $x$ .

If the denominator contains a square term $x^2$ or a square root term $\sqrt{x}$ we can also apply the same strategy. Here is an example:

Example 5

Solve the equation

\frac{2}{3x^2}-5=0

We solve as usual, but multiply each side by $x^2$ :

\begin{array}{llll} \frac{2}{3x^2}-5 &=& 0 &\quad| +5\\ \frac{2}{3x^2} &=& 5 &\quad| \cdot x^2\\ \frac{2}{3} &=& 5x^2 &\quad| :5\\ \frac{2}{15} &=& x^2 &\quad| \sqrt{\phantom{a}}\\ \pm \sqrt{\frac{2}{15}} &=& x \end{array}

Thus, the two solutions are $x_1=-\sqrt{\frac{2}{15}}=-0.365$ and $x_2=\sqrt{\frac{2}{15}}=0.365$

Equations with a product that results in zero

Recall that factors are the terms involved in a multiplication. For example, the multiplication

3\cdot 5

has the factors $3$ and $5$ , and the multiplication

(x-1)x(x^2+2)

has the factors $x-1$ , $x$ , and $x^2+2$ . We discuss now how to solve equations which consists of factors whose product equals zero, the simplest case being

\boxed{(x-a)(x-b)=0}

An example is

(x-1)(x+2)=0

To solve this equation, we start with a simple observation: For any two numbers, say $a\cdot b$ , we know that the product cannot be zero, unless $a$ or $b$ are zero (or both). Indeed, a product like $3\cdot 5=15\neq 0$ , but $0\cdot 5=0$ and $3\cdot 0=0$ .

Let's apply this observation to our equation, say,

(x-1)(x+2)=0

As it is a product of two factors, $x-1$ and $x+2$ , it can only be zero if the factor $x-1$ is zero, or the factor $x+2$ is zero. And it is simple to find these values:

\begin{array}{ll} x-1=0 & \rightarrow x=1\\ x+2=0 & \rightarrow x=-2\\ \end{array}

So $1$ and $-2$ must be the solutions of this equation. Let's check. For $x=1$ we have

(1-1)(1+2)=0\cdot 3 = 0

so $x=1$ is a solution. For $x=-2$ we get

(-2-1)(-2+2)=-3\cdot 0=0

so $x=-2$ is a solution as well.

It is straightforward to extend this method to equations consisting of more factors, which can also be more complex. Consider the following example:

Example 6

Solve the equation

(x-1)x(x^2-2)=0

According to our method, we have to find all $x$ -values such that the factor $x-1$ is zero (thus $x=1$ is a solution), or the factor $x$ is zero (thus $x$ =0 is a solution), or the factor $x^2-2$ is zero. For this last factor, we have to find an $x$ with

\begin{array}{llll} x^2 -2 & = & 0 & \quad| +2\\ x^2 & = & 2 & \quad| \sqrt{\phantom{a}}\\ x &=& \pm \sqrt{2} \end{array}

So the solutions are $x_1=1,x_2=0,x_3=\sqrt{2}$ and $x_4=-\sqrt{2}$ .

Sometimes we first have to bring an equation into the right form, that is, write it as factors. Mostly, this is done by factoring out. Here is an example:

Example 7

Solve the equation

2x^2-3x=0

Let's factor out the $x$ , which appears in both terms:

\begin{array}{lll} 2x^2-3x & = & 0 & |\quad \text{factor out}\\ x(2x-3) & = & 0 \end{array}

The first factor is $x$ , so one solution is $x=0$ . The other factor is $2x-3$ . To find out the $x$ -value for which is factor is zero, we have to solve a linear equation:

\begin{array}{llll} 2x-3 & = & 0 & |\quad +3\\ 2x & = & 3 & |\quad :2 \\ x &=& 1.5 \end{array}

So the solutions are $x_1=0$ and $x_2=1.5$ .

Exercise 1

Solve the following equations:

$x^2-3=5$
$2x^2-4=6x^2-10$
$3\sqrt{x}=4$
$\sqrt{2x}-2=4$
$4x^2-4 = x^2+8$
$3\sqrt{x}-2=0.5 \sqrt{x}$
$\frac{2}{x}=4$
$\frac{2}{x}-1=4$
$\frac{2}{\sqrt{x}}= 5$
$x(x+1)=0$
$x(x+1)(x+2)=0$
$\frac{4}{5x^2} = \frac{2}{10x^2}+1$
$x(2x+1)(x^2-1)=0$
$x^2-x=0$
$3x^3-2x^2=0$
$4x^3-x=0$
$(x-1)x(x^2+2)=0$
$5(x-1)^2=0$
$x^2(x-1)=0$
$3(x-1)(x-2)(x-3)(x-4)=0$

Solution

$x_{1,2}=\pm \sqrt{8}=\pm 2.828$
$x_{1,2}=\pm \sqrt{1.5}=\pm 1.225$
$x = \frac{16}{9}=1.778$
$x=18$
$x_{1,2}=\pm 2$
$x=\frac{16}{25}=0.64$
$x=\frac{1}{2}=0.5$
$x=\frac{2}{5}=0.4$
$x=\frac{4}{25}=0.16$
$x_1=0, x_2=-1$
$x_1=0, x_2=-1, x_3=-2$
$x_{1,2}=\pm\sqrt{\frac{3}{5}}=\pm 0.775$
$x_1=0, x_2=-0.5, x_{3,4}=\pm 1$
$x_1=0, x_2=1$
$x_1=0, x_2=\frac{2}{3}=0.\overline{6}$
$x_1=0, x_{2,3}=\pm\frac{1}{2}$
$x_1=1, x_2=0$
$x=1$
$x_1=0, x_2=1$
$x_1=1, x_2=2, x_3=3, x_4=4$