An area that must be learned in mathematics is quadratic equations. Compared to linear equations, quadratic equations are more complex to calculate.

In quadratic equations, there are multiple ways to answer the question. Square roots, factorization, and quadratic formulas are the most common ways to solve quadratic equations. When solving a quadratic equation, we must distinguish which method is the best way to solve it. This is why quadratic equations tend to be more difficult to solve than linear equations.

We can answer quadratic equations by using what we have learned so far, such as square roots and factorization. Also, since word problems of quadratic equations are frequently given, we must be able to make equations.

We will explain how to solve quadratic equations, including not only how to answer the questions, but also how to solve word problems.

Table of Contents

## An Equation Containing $x^2$ Is a Quadratic Equation

What is a quadratic equation? An expression that is connected to two expressions by equals sign is an equation. An equation is called a quadratic equation if $x^2$ exists in the equation.

Therefore, the following are all quadratic equations.

- $x^2=9$
- $x^2+4x+3=0$
- $(x+5)^2=36$

Every expression has $x^2$. Such an equation is a quadratic equation. There is a major difference between linear and quadratic equations. That is, there are two answers in a quadratic equation. For example, what is the answer to the following quadratic equation?

- $x^2=16$

As for the number that squared results in 16, there is 4. 4^{2} is 16. Therefore, $x=4$ is the answer. However, squaring -4 will still result in 16. The answer to (-4)^{2} is 16. Therefore, $x=-4$ is also the answer.

It is important to understand that in quadratic equations, there can be two answers, not just one. Sometimes there is one answer, and sometimes there are two. This is the main difference between linear and quadratic equations.

## Quadratic Equations Have Three Ways to Solve

In quadratic equations taught in mathematics, there are three ways to solve them. To be more specific, the following three methods are used.

- Using the square root.
- Utilize the factorization.
- Utilize the quadratic formula.

We can use any one of these methods to answer the question. Different quadratic equations give different methods of answering the question, so we should be able to distinguish which method is the best.

### Solving Quadratic Equations Utilizing Square Root

When answering a quadratic equation, there is a way to solve it using the square root. There are not many situations in which square roots can be used. However, when they are available, square roots make it easy to find answers.

To get answers by square root, the answer must be in the form of a quadratic equation, such as the following.

- $ax^2=b$
- $(x+a)^2=b$

If we can connect the equation of squares with equals sign, we can use the square root. The way to answer the question is to simply add a root symbol. We can eliminate the squares by adding the radical symbol, as shown below.

When solving quadratic equations, the easiest way to answer a problem is through the use of square roots. We can answer a quadratic equation simply by adding a radical sign.

### Factor the Left Side and Set the Right Side to 0

When answering a quadratic equation, a frequently used method is factorization. The use of factoring formulas is the way to get an answer.

There are four factoring formulas.

- $(x+a)(x+b)=x^2+(a+b)x+ab$
- $(x+a)^2=x^2+2ax+a^2$
- $(x-a)^2=x^2-2ax+a^2$
- $(x+a)(x-a)=x^2-a^2$

If you do not remember this formula, you will not be able to factor it. That is, you will not be able to solve problems with quadratic equations. Remembering factoring formulas is essential for solving math problems.

So how do we factor it? When answering a quadratic equation by factorization, make sure $A=0$. The right side must be zero. For example, what is the answer to the following quadratic equation?

- $x^2+6x+5=0$

First, let’s factor this equation. Using the factoring formula, we can factor it into the following equation.

- $(x+1)(x+5)=0$

We need to answer the number of $x$ that satisfies this quadratic equation. What number can we substitute into $x$ that will result in zero?

In mathematics, any number has the property that if we multiply it by zero, the answer will be zero. So, if $(x+1)$ or $(x+5)$ is zero, the answer will be zero and match the right side.

As the number of $x$ for which $(x+1)$ or $(x+5)$ is zero, it is -1 and -5. In other words, $x=-1,x=-5$ is the correct answer. Thus, after setting the right side to 0, we factorize the left side. Then we can get the answer to the quadratic equation.

### Get a Complicated Answer by the Quadratic Formula

However, in some cases, the answer can be complicated.

If we use square roots, we must make a squared expression. If we use factorization, we must have a factorization of the left side. So how do we find the answer to a quadratic equation if neither square root nor factorization can be used?

In this case, we use the quadratic formula. There is a formula for solving a quadratic equation. If $ax^2+bx+c= 0$, the formula that gives the answer to $x$ is the quadratic formula. The quadratic formula is as follows.

$$x={-b\pm\sqrt{b^2-4ac}\over2a}$$

It is a quite complicated formula and this is the quadratic formula. We can check the quadratic equation and substitute the numbers into the quadratic formula to get the answer. For example, consider the following quadratic equation.

- $2x^2+3x-4=0$

If we want to get an answer to this quadratic equation, we need to substitute the following to the quadratic formula.

Even if we can’t use square roots or factorization, we can still get the answer by using the quadratic formula. Although the quadratic formula must be memorized, the answer can be obtained by substituting numbers.

It is possible to explain why the quadratic formula holds, using our previous knowledge of mathematics.

However, the formula for proving it is complex. Also, no one proves the formula every time they use the quadratic formula to come up with an answer. Therefore, we will skip explaining why the quadratic formula is valid. The quadratic formula should be memorized.

### Use a Simplified Quadratic Formula If $b$ Is Even

Note that when using quadratic formulas to solve problems, it is common to use a simplified quadratic formula if $b$ is even.

For even numbers, we can always divide by 2. So, set $b’$ after dividing $b$ by 2. That is, $b’=\displaystyle\frac{b}{2}$. In that case, the following quadratic formula holds.

$$x={-b’\pm\sqrt{b’^2-ac}\over a}$$

The meaning of the quadratic formula is exactly the same, so we don’t always have to remember this formula. However, it is recommended to keep this simplified formula in mind when solving problems in mathematics. Because it will speed up your calculations and also prevent miscalculation.

For example, try to solve the following quadratic equation using the quadratic formula.

- $2x^2+4x-1=0$

A comparison of each solution is as follows.

Both answers are the same. However, if we do not use the simplified quadratic formula, the calculation becomes more complicated. By doing prime factorization, we have to reduce the number in the root symbol. We also need to reduce the numbers in the fraction by dividing it.

In mathematics, the larger the number and the more steps required, the more likely it is that calculation mistakes will occur. To prevent miscalculation, most people use a simplified quadratic formula when answering quadratic equations.

## How to Solve Word Problems of Quadratic Equations

In this way, try to determine which of the three methods of solving the problem is best.

In addition, in mathematics, the most important thing is to use it in real life. Not only we can calculate it, but we should actually use it. For this reason, we need to be able to solve word problems. The way to solve a word problem is the same as for linear and simultaneous equations. The steps are as follows.

- Make $x$ for the unknown number.
- Make quadratic equations from the problem statement.
- Solve quadratic equations.

So far, if you have solved word problems in mathematics, the method is the same for quadratic equations.

### Be Sure to Check If the Answer Is Optimal

However, in the case of quadratic equations, we must pay attention. That is, we must always make sure that the answer is optimal.

As mentioned earlier, quadratic equations often give two answers. Then, one answer is frequently not optimal. Although the answer is available as a number, it is often not possible in the real world.

For example, let’s solve the following quadratic equation.

- The area of a rectangle with a height $x$ cm and a width $(x-2)$ cm is 24 cm
^{2}

The area of a rectangle can be found by the formula “vertical × horizontal.” Therefore, we can make the following quadratic equation.

- $x(x-2)=24$

Solving this quadratic equation gives us the following.

$x(x-2)=24$

$x^2-2x=24$

$x^2-2x-24=0$ (to factorize)

$(x-6)(x+4)=0$

$x=-4,x=6$

As you can see, solving a quadratic equation yields two answers. However, $x=-4$ is not the answer. If $x=-4$, then the equation $x(x-2)=24$ is satisfied. However, in the real world, the vertical length of a rectangle cannot be -4 cm. Therefore, the answer is only 6 cm.

If it is a temperature problem, the answer can be negative. In the real world, it is normal for the temperature to be negative. However, the length of the rectangle’s vertical and horizontal sides should not be negative.

One thing to keep in mind with word problems in quadratic equations is that the answer we get is not always the correct answer. Even if we come up with an answer for $x$, be sure to check if the answer is suitable.

## Exercise: Exercise: Quadratic Equation Calculations and Word Problems

**Q1.** Solve the following quadratic equation.

- $2x^2-12=0$
- $(x+2)(x+4)=24$
- $x^2+5x+2=0$

**A1.** Answers.

There are three ways to solve quadratic equations. Make sure which method is the best for you to solve the problem.

**(a)**

Solve using square roots.

$2x^2-12=0$

$2x^2=12$

$x^2=6$

$x=\sqrt{6},x=-\sqrt{6}$

**(b)**

Solve using factorization.

$(x+2)(x+4)=24$

$x^2+6x+8=24$

$x^2+6x-16=0$

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

$x=2,x=-8$

**(c)**

Solve using the quadratic formula.

$x^2+5x+2=0$

$\displaystyle x={-5\pm\sqrt{5^2-4×1×2}\over 2×1}$

$\displaystyle x={-5\pm\sqrt{25-8}\over 2}$

$\displaystyle x={-5\pm\sqrt{17}\over 2}$

**Q2.** Solve the following word problem.

There is a rectangle land with a width 4 m longer than the vertical. On this land, make a path of 1 m wide and create four flowerbeds as follows.

If the area of all flowerbeds is 45 m^{2}, how much is the length of the vertical?

**A2.** Answer.

For example, if you are working in architectural design, you have to do this kind of calculation. Mathematics is meant to be utilized in practice, and quadratic functions are used in many situations.

When solving this word problem, write the numbers in the figure. Otherwise, the problem will be difficult to solve. If the length of the vertical is $x$ m, we have the following.

It is also difficult to understand if the four flowerbeds exist separately. So, let’s move all the flowerbeds to the top left, hypothetically.

The area of all flowerbeds will not change. In other words, moving the flowerbeds does not change the number of the answer. When we actually build the flowerbed, we can return it to its original location. So, let’s make a quadratic equation considering the hypothetical situation where it has been moved. Then we have the following.

Since one path is 1 meter wide, the total width of the three paths is 3 meters. Therefore, the total vertical length of the flowerbed is $(x-3)$ m. And the total horizontal length of the flower bed is $(x+1)$ m.

Since the area of all flowerbeds is 45 m^{2}, we can make the following quadratic equation.

- $(x-3)(x+1)=45$

Let’s solve this quadratic equation.

$(x-3)(x+1)=45$

$x^2-2x-3=45$

$x^2-2x-3-45=0$

$x^2-2x-48=0$

$(x-8)(x+6)=0$

$x=8,x=-6$

By solving the quadratic equation, we get the answer $x=8,x=-6$. However, the length of the vertical of the land cannot be -6 m. The number must be a positive number. Therefore, -6 m is unsuitable as an answer and must be excluded. Therefore, the correct answer is a vertical length of land is 8 meters.

In math problems, it is often very difficult to solve the problems by looking at the figures in the problem statement. However, as in this exercise, moving the figure around often simplifies the problem and allows us to create quadratic equations.

As long as the numbers are correct, you can transform the figure as you wish. Try to think about how you can change the diagram to make it easier to calculate.

## Solving Quadratic Equations Calculating Problems

An expression with $x^2$ is a quadratic equation. Solving quadratic equations in mathematics increases the difficulty. It requires us to use multiple pieces of knowledge that we have learned previously.

When solving quadratic equations, we must understand the following.

- Square roots
- Factorization

We also need to memorize the quadratic formula. There are three ways to solve a quadratic equation, and we have to distinguish which method is the best. In addition, there are two quadratic formulas, so we should be able to use both.

Furthermore, in quadratic equation word problems, one of the answers may be unsuitable. Compared to linear equations, quadratic equations require more attention and are more complex to solve. Let’s understand how to solve quadratic equations.