# Solving Quadratic Equations – Methods and Examples

Quadratic equations have the form ax²+bx+c=0. These equations can be solved using various methods depending on the type of quadratic equation we have. We can use methods for incomplete equations, solve equations by factoring, by completing the square, or with the quadratic formula.

Here, we will learn how to solve quadratic equations using different methods. We will use several examples to improve the learning process of the methods.

##### ALGEBRA

Relevant for

Learning to solve quadratic equations with various methods.

See methods

##### ALGEBRA

Relevant for

Learning to solve quadratic equations with various methods.

See methods

An incomplete quadratic equation is an equation that does not have a term from the form $latex ax^2+bx+c=0$, as long as the x² term is always present. When this is the case, we have two types of incomplete quadratic equations depending on the missing term.

### Solving quadratic equations that do not have the term bx

To solve quadratic equations of the form $latex ax^2+c=0$ that do not have the bx term, we need to isolate x² and take the square root of both sides of the equation.

For example, suppose we want to solve the equation $latex x^2-9=0$. First, we have to write it as follows:

$latex x^2=9$

Now that we have x² on the left-hand side, we can take the square root of both sides of the equation:

$latex x=\sqrt{9}$

$latex x=\pm 3$

Note: We must consider both the positive solution and the negative solution, since $latex (-3)^2=9$.

### Solving quadratic equations that do not have the term c

To solve equations of the form $latex ax^2+bx=0$ that do not have the constant term c, we have to factor the x on the left-hand side of the equation. Then, we form two equations with the factors and solve them.

For example, suppose we want to solve the equation $latex x^2-5x=0$. First, we factor it as follows:

$latex x(x-5)=0$

Since we have two factors, we can form an equation with each factor and solve:

$latex x=0~~$ or $latex ~~x-5=0$

$latex x=0~~$ or $latex ~~ x=5$

Note: In this type of equation, one of the solutions will always be $latex x=0$.

## Solving quadratic equations by factoring

The factorization method consists of finding the factors of the quadratic equation so that we have the roots of the equation exposed. By forming an equation with each factor, we can find the roots.

We can solve quadratic equations by factoring by following these steps:

Step 1: Simplify the equation if possible and write it in the form $latex ax^2+bx+c=0$.

Step 2: Find the factors of the equation using any method and write it in the form $latex (x+p)(x+q)=0$.

Step 3: Take each factor and set it equal to zero to form an equation. For example, $latex x+p=0$.

Step 4: Solve the equation for each factor.

There are several methods we can use to factor quadratic equations. The general idea is to find two factors of the form $latex (x+p)(x+q)=0$, which result in the form $latex x^2+bx+c=0$ when multiplied.

For example, the equation $latex x^2+2x-3=0$ can be factored into the form $latex (x+3)(x-2)=0$, since multiplying the factors gives us the original equation.

You can learn or review how to factor quadratic equations by visiting our article: Factoring Quadratic Equations.

This method allows us to find both roots of the equation relatively easily. However, it is not always possible to factor a quadratic equation.

## Solving quadratic equations by completing the square

Completing the square is a factoring technique that allows us to write an equation from the form $latex ax^2+bx+c=0$ to the form $latex (x-h)^2+k=0$. Thus, we can solve quadratic equations that cannot be easily factored.

To solve quadratic equations using the method of completing the square, we can follow the steps below:

Step 1: Simplify and write the equation in the form $latex ax^2+bx+c=0$.

Step 2: When a is different from 1, the entire equation must be divided by a to obtain an equation with a value of a equal to 1:

$latex x^2+bx+c=0$

Step 3: Divide the coefficient b by 2 to obtain:

$$\left(\frac{b}{2}\right)$$

Step 4: Square the expression from step 2:

$$\left(\frac{b}{2}\right)^2$$

Step 5: Add and subtract the expression obtained in step 4 to the equation obtained in step 2:

$$x^2+bx+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c=0$$

Step 6: Factor the equation using the identity $latex x^2+2xy+y^2=(x+y)^2$:

$$\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c=0$$

Step 7: Simplify to obtain an equation of the following form:

$latex (x-h)^2+k=0$

Step 8: Rearrange the equation as follows:

$latex (x-h)^2=-k$

Step 9: Take the square root of both sides of the equation:

$latex x-h=\sqrt{-k}$

Step 10: Solve for x:

$latex x=h\pm \sqrt{-k}$

When it is not possible to solve quadratic equations with any other method, we can use the quadratic formula, since this method allows us to find both solutions of any quadratic equation.

To use the general quadratic formula, we have to write the equation in the form $latex a{{x}^2}+bx+c=0$. This will allow us to identify the values of the coefficients a, b and c easily. We then use those values in the quadratic formula:

Note: We must not forget the ± sign, since in this way we will obtain both solutions to the quadratic equation when it is the case.

The expression inside the square root of the quadratic formula ($latex b^2-4ac$) is the discriminant of the quadratic equation. The discriminant determines the type of root that the quadratic equation will have.

Therefore, depending on the value of the discriminant, we have the following:

• When $latex b^2-4ac>0$, the equation has two real roots.
• When $latex b^2-4ac<0$, the equation has no real roots.
• When $latex b^2-4ac=0$, the equation has a repeated root.

When the value inside the square root of the formula is positive, we will have two real roots. When that value is negative, we won’t have real roots (but we will have imaginary or complex roots). When that value is equal to zero, we will have a single root.

The following examples are solved using all the methods for solving quadratic equations studied above. Try to solve the problems yourself before looking at the solution.

### EXAMPLE 1

What are the solutions of the equation $latex x^2-4=0$?

This equation is an incomplete quadratic equation that does not have the bx term. Therefore, we can find the solutions by isolating the quadratic term and taking the square root of both sides of the equation:

$latex x^2-4=0$

$latex x^2=4$

$latex x=\pm\sqrt{4}$

$latex x=\pm 2$

The solutions of the equation are $latex x=2$ and $latex x=-2$.

### EXAMPLE 2

Solve the equation $latex x^2-7x=0$.

This equation is an incomplete quadratic equation that does not have the constant term c. We can solve it by factoring the x and forming an equation with each factor:

$latex x^2-7x=0$

$latex x(x-7)=0$

$latex x=0 ~~$ or $latex ~~x-7=0$

$latex x=0 ~~$ or $latex ~~x=7$

The solutions of the equation are $latex x=0$ and $latex x=-7$.

### EXAMPLE 3

Solve the equation $latex x^2+2x-8=0$ using the factoring method.

Factoring the left-hand side of the equation, we have:

$latex x^2+2x-8=0$

$latex (x+4)(x-2)=0$

Now, we form an equation with each factor and solve:

$latex x+4=0~~$ or $latex ~~x-2=0$

$latex x=-4~~$ or $latex ~~x=2$

The solutions of the equation are $latex x=-4$ and $latex x=2$.

### EXAMPLE 4

Solve the equation $latex 2x^2-13x-24=0$ using the factoring method.

We can factor the left-hand side of the equation as follows:

$latex 2x^2-13x-24=0$

$latex (2x+3)(x-8)=0$

Now, we form an equation with each factor and solve:

$latex 2x+3=0~~$ or $latex ~~x-8=0$

$latex x=-\frac{3}{2}~~$ or $latex ~~x=8$

The solutions of the equation are $latex x=-\frac{3}{2}$ and $latex x=8$.

### EXAMPLE 5

Use the method of completing the square to solve the equation $latex x^2+4x-6=0$.

In this equation, the coefficient b is equal to 4. Therefore, we have:

$$\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2$$

$$=2^2$$

Now, we have to add and subtract that value to the quadratic equation:

$$x^2+4x-6=x^2+4x+2^2-2^2-6$$

Completing the square and simplifying, we have:

$latex = (x+2)^2-4-6$

$latex = (x+2)^2-10$

Rearranging, we form the equation:

$latex (x+2)^2=10$

Taking the square root of both sides, we have:

⇒ $latex x+2=\sqrt{10}$

Solving, we have:

⇒ $latex x=-2\pm \sqrt{10}$

### EXAMPLE 6

Solve the equation $latex 2x^2+8x-10=0$ by completing the square.

We can divide the entire equation by 2 to make the coefficient of the quadratic term equal to 1:

⇒ $latex x^2+4x-5=0$

Now, we can see that the coefficient b is equal to 4. Therefore, we have:

$$\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2$$

$$=2^2$$

If we add and subtract that expression to the quadratic equation, we have:

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

Completing the square and simplifying, we have:

$latex = (x+2)^2-4-5$

$latex = (x+2)^2-9$

Now, we rearrange the equation as follows:

⇒ $latex (x+2)^2=9$

And we take the square root of both sides:

⇒ $latex x+2=3~~$ or $latex ~~x+2=-3$

Solving, we have:

⇒ $latex x=1~~$ or $latex ~~x=-5$

### EXAMPLE 7

Use the general formula to solve the equation $latex 2x^2+3x-4=0$. Express the solutions to two decimal places.

This equation has the coefficients $latex a=2$, $latex b=3$, and $latex c=-4$. Therefore, using the quadratic formula with those values, we have:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(3)\pm \sqrt{( 3)^2-4(2)(-4)}}{2(2)}$$

$$=\frac{-3\pm \sqrt{9+32}}{4}$$

$$=\frac{-3\pm \sqrt{41}}{2}$$

$$=-2.35 \text{ or }0.85$$

The solutions of the equation are $latex x=-2.35$ and $latex x=0.85$.

## Solving quadratic equations – Practice problems

Solve the following problems using any of the methods we studied above.