**The quadratic equations**

*Quadratic equations* are the type of equations in the form

$$ax^2 + bx + c = 0,$$

where $a, b, c \in \mathbb{R}$ are and $a\neq 0$.

If $a = 0$, our equation reduces to $bx + c = 0$ which is linear equation. More about solving linear equations you can find in lessons One-step equations, Two-step equations and Multi-step equations.

Number $a$ in this equation is called the *leading coefficient*, number $b$ is the *linear coefficient *and $c$ is *constant*.

Every $x$ (real or complex) that satisfies this equation is called the *solution of the quadratic equation*.

**The formula for the solutions of the quadratic equation**

The universal method of solving these kinds of quadratic equations is by using the formula for the solutions of the quadratic equation:

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}.$$

**Example** Solve the equation $\ x^2 + 5x + 4 = 0$.

First we determine coefficients: $\ a = 1$, $\ b = 5$ and $\ c = 4$. Now we simply insert coefficients into the formula for the solutions of the quadratic equation:

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} = \frac {-5 \pm \sqrt{5^2 – 4\cdot\ 1 \cdot 4}}{2 \cdot 1} = \frac{-5 \pm 3}{2}$$

$$\Rightarrow x_1 = \frac{-5 + 3}{2} = -1 \quad \text{and} \quad x_2 = \frac{-5 – 3}{2} = -4.$$

We can check if result is correct by inserting the solutions we obtained into the given equation:

$$x_1^2 + 5x_1 + 4 = (-1)^2 + 5(-1) + 4 =1-5+4=0$$

and

$$x_2^2 + 5x_2 + 4= (-4)^2 + 5(-4) + 4=16-20+4 =0.$$

**Special forms of quadratic equations**

If quadratic equation is in special form sometimes it is easier to manipulate given equation to find solutions instead of using formula for solutions of quadratic equations. Some of these special forms are:

1) If $\ b = 0$ and $\ c \neq 0$ then the equation is:

$$ax^2 + c = 0.$$

First, from both sides of equation we substract

$$ax^2=-c$$

and then divide the entire equation by $a$

$$x^2 = – \frac {c}{a}.$$

This equation always has two different solutions. The solutions may be real or complex numbers, depending on the sign of numbers $a$ and $c$.

If $- \frac {c}{a}>0$ then the quadratic equation has two real solutions:

$$x_1 = \sqrt{\left(-\frac{c}{a}\right)} \text{ and } x_2 =-\sqrt{\left(-\frac{c}{a}\right)}.$$

If $-\frac{c}{a} < 0$, then the solutions are complex numbers

$$x_1 = i \sqrt{\left|-\frac{c}{a}\right|} \text{ and } x_2 = – i \sqrt{\left|-\frac{c}{a}\right|}.$$

** Example** Solve the equation $\ 2x^2 – 8 = 0$.

We add $8$ to both sides of equations and obtain $ 2x^2 = 8$. Now we divide equation with 2 to get $x^2 = 4$. Because $2^2=4$ and $(-2)^2=4$ solutions are $x_1 = 2$ and $x_2=-2$, that is $x_{1,2}=\pm\sqrt{4}.$

**Example** Solve the equation $\ 2x^2 + 8 = 0$.

The procedure is similar to the one in example above, we substract 8 from both sides and divide equation by 2 to obtain $\ x= \sqrt{(-4)}$. This means that solutions of this quadratic equation are not real numbers, they are complex: $\ x_1 = -2i$, $\ x_2 = 2i$.

2) $\ c = 0$ and $b \neq 0$.

The quadratic equation is now given in the form:

$$\ ax^2 + bx = 0.$$

We can extract $x$ from both terms and write equation as $x(ax+b)=0$. Now notice that if $x=0$ or $ax+b=0$ left side of equation is equal to zero. So $x=0$ is one solution and other solution is solution of linear equation $ax+b=0$. Solution of that linear equation is $x=\frac{-b}{a}$ (we substracted $b$ from both sides and divided it by $a$).

**Example** Solve the equation $\ 5x^2 + 4x = 0$.

First, we extract $x$. Then we have $\ x(5x + 4) = 0$. One solution is $x=0$. We find second solution by solving $\ 5x + 4=0$:

$$\ 5x + 4 = 0$$

$$\ 5x = – 4 $$

$$\ x = – \frac{5}{4}.$$

3) If $b = 0$ and $c = 0$ equaion reduces to $ax^2 = 0$ and we immediately see that only solution is $x=0$.

$x = 0$ is called a *double solution *of the quadratic equation $ax^2 = 0$.

**Factoring**

Other way of simplifying is factoring. If we have two factors whose product is equal to zero we can easily find solutions. Suppose we have written quadratic equation $ax^2+b+c=0$ as $k(x-p)(x-q)=0$ for some real numbers $k, p$ and $q$. Then obviously $p$ and $q$ are solutions of original equation (if we insert $p$ or $q$ in given equation we see that equality holds).

**Example **Solve the equation $\ x^2 + 6x + 9 = 0$ by factoring.

Using the formula for square of sum (see lesson Determining polynomials, basic math operations, the most important rules for multiplying, under section Multiplication) we write $x^2 + 6x + 9$ as $(x+3)^2$ which can be written as $(x+3)(x+3)$ so given equation is equivalent to $(x+3)(x+3)=0$. We now see that $x=3$ is double solution.

**Solving quadratic equations by completing the square**

Useful technique for solving quadratic equations is completeng the square. We are going to tranform our original equation $ax^2+bx+c=0$ to more appropriate form from which we can easier find solutions.

First, we extract leading coefficient $a$:

$$ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})$$

Now we add and substract $\left(\frac{b}{2a}\right)^2$:

$$a(x^2+\frac{b}{a}x+\frac{c}{a})=a(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a})$$

and group terms in a way that allows us to use square of sum formula

$$a(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a})$$

$$=a((x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2)-\left(\frac{b}{2a}\right)^2+\frac{c}{a})$$

$$=a((x+\frac{b}{2a})^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a})=a(x+\frac{b}{2a})^2-a\left(\frac{b}{2a}\right)^2+a\frac{c}{a})$$

$$=a(x+\frac{b}{2a})^2-a\frac{b^2}{4a^2}+c=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c.$$

So we have $ax^2+bx+c=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$.

Algorithm above, with geometric interpretation is shown in animation below.

To find solution of $ax^2+bx+c=0$ we can find solutions of $a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c=0$. We can substitute $y=x+\frac{b}{2a}$ and $d=-\frac{b^2}{4a}+c$ and equation transforms to $ay^2+d=0$ which we know how to solve from the first case of special form equations.

To understand better the process we just decribed look at the next example.

*Example 6.*

If we have a term such as $\ 4x^2 + 4x + 5$ we can express it as $\ (2x)^2 + 2\cdot 2x + 1 + 4$.

We have a first member squared, second doubled and third squared.

We can write that down as $\ (2x+1)^2 + 4$.

*Example 7.*

We are observing the quadratic equation $8x^2 + 6x + 2 = 0$.

Let’s write the equation now in form:

From the left and right side of the equation we previously subtracted $-2$. We apply a methodology of completing the square on the left side of the equation.

$8$ is not a whole square, but $4$ is. This means that we can first extract $2$ and get:

Now we have the first member squared. It can be observed that the square of the second member is missing, which is $3$, because the first member when multiplied by $3$ is equal to $6x$. Also, we need to multiply $6x$ by $2$ because the first member was multiplied by $2$.

Therefore, now we have:

$$2 \cdot (2x)^2 + 2 \cdot 3x + 2 \cdot 3^2 = -2 +2\cdot 3^2 $$

$$ 2 \cdot(2x + 3)^2 = 16$$

$$ (2x + 3)^2 = 8.$$

How can this help us solve quadratic equations?

*Example 8.* Solve the equation $\ 8x^2 + 6x + 2 = 0$ by completing the square.

*Solution*:

From the previous example above it was expressed that the quadratic equation $\ 8x^2 + 6x + 2 = 0$ can be written as:

$$ (2x + 3)^2 = 8.$$

$\ 2x = -3 \pm \sqrt{8}$

$\ x = \frac{-3 \pm \sqrt{8}}{2}$.

## The discriminant of the quadratic equation

** The** **discriminant** of the quadratic equation $ax^2 + bx + c$ is the number $D$:

$$D=b^2 – 4ac.$$

Then the solutions of the quadratic equation written by using the discriminant are:

$$x_{1,2}= \frac{-b \pm \sqrt{D}}{2a}.$$

If $\ D > 0$, then the quadratic equation thus has two distinct real solutions, if $\ D < 0$, the quadratic equation has two solutions that are complex conjugates , and if $\ D = 0$ the quadratic equation has one real solution of multiplicity two.

## Vieta’s formulas

French mathematician François Viète also studied quadratic equation and came to an important relation between quadratic equation and system of two equations with two unknowns. Those unknowns are the solutions to observed quadratic equation.

The solutions $x_1$ and $x_2$ of the quadratic equation $\ ax^2 + bx + c = 0$ meet the criteria of the **Vieta’s formulas:**

$$x_1 + x_2 = -\frac{b}{a},$$

$$x_1 \cdot x_2 = \frac{c}{a}.$$

*Example 9.*

If $\ x = 3$ is one solution to the equation $\ x^2 + bx + 8 = 0$, what is the other solution and what is $b$?

*Solution*:

According to the Vieta’s formulas we find that:

$$x_1 + x_2 = -\frac{b}{1}$$

$$x_1 \cdot x_2 = \frac{8}{1}.$$

Since we already have one solution this can be written as:

$$ 2 + x_2 = -b$$

$$2x_2 = 8.$$

From the second equation we find that $\ x_2 = 4$, which leads us to $\ b = -6$.

*Example 10.*

Determine the product and sum of the solutions of equation $\ x^2 + 2x – 5 = 0$.

*Solution:*

From the equation we read coefficients:

$$a = 1, b = 2, c = -5 $$.

Now, by using Vieta’s formulas we have:

$$ x_1 + x_2 = – \frac{b}{a}= -2, \quad x_1\cdot x_2 = \frac{c}{a} = -5 $$.

## The system of linear and quadratic equations

The quadratic equation with two unknowns has the form:

where $A, B, C, D, E, F$ are real numbers – coefficients.

This kind of equation cannot be solved without any terms attached to it, however, if we have one more additional condition like a linear equation, it can be solved. This is because from the linear equation we get the information in which relation are unknowns $x$ and $y$ and we can then extract one using the other to get the quadratic equation with only one unknown. The solutions to this system are two ordered number pairs $(x_1, y_1)$ and $(x_2, y_2)$.

*Example 11.*

Solve the following system of linear and quadratic equation:

*Solution*:

First we will observe a linear equation and extract one unknown according to the other one. It doesn’t matter which unknown we choose. In this case, we choose $y$:

$\ x^2 + 4x – 2x^2 – 6(4 – 4x + x^2) = 0$

$\ -x^2 + 4x – 24 + 24x – 6x^2 = 0$

$\ -7x^2 + 28x – 24 = 0$

$\ 7x^2 – 28x + 24 = 0$.

The solutions of the quadratic equation $7x^2 – 28x +24 = 0$ are:

$$ x_{1,2} = \frac{28 \pm \sqrt{28^2 – 4 \cdot 7 \cdot 24}}{2\cdot 7} = \frac{28 \pm \sqrt{784-672}}{14} =\frac{28 \pm \ 4\sqrt{7}}{14} = 2 \pm \frac{2\sqrt{7}}{7}$$.

Now we must solve linear equations $y_1 = 2 – x_1$ and $y_2 = 2 – x_2$.

As follows:

$$y_1 = 2 – 2 – \frac{2\sqrt{7}}{7} = – \frac{2\sqrt{7}}{7},$$

$$y_2 = 2 – 2 + \frac{2\sqrt{7}}{7} = \frac{2\sqrt{7}}{7}.$$

The solutions are: $(2+\frac{2\sqrt{7}}{7}, – \frac{2\sqrt{7}}{7})$, and $(2-\frac{2\sqrt{7}}{7}, \frac{2\sqrt{7}}{7})$.

## Biquadratic equations

**Biquadratic**** equations** are equations with two unknowns to the power of two. They are solved by using substitution.

*Example 12.* Solve the following system:

$\ x^2 – 4y^2 = -3.$

First we would like to have only one unknown in our equation so we will extract by $y =\frac{1}{x}$ and insert it in other equation:

$$ x^4 – 4 + 3x^2 = 0$$

$$x^4 + 3x^2 – 4 = 0.$$

Now we can make a substitution. We will remove of $x$ to the power of $4$ through substitution $\ x^2$ with $u$ to get the quadratic equation:

$$\Rightarrow \ u_1 = 1 , u_2 = -4$$.

We are still not finished, we have to return to our substitution and solve the quadratic equation:

For $ u_1$ we have:

$$x^2 = 1 \Rightarrow x_1 = 1, \quad x_2 = -1$$.

For $ u_2$ we have:

$$ x^2 = -4 \Rightarrow x_3 = -2i, \quad x_4 = 2i$$.

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