The quadratic equation

Although this method will not work for all quadratic equations, it is always worth checking because it can provide us with a quick solution without needing to solve the quadratic formula first.

It is important to note that, unlike the quadratic formula, solving a quadratic equation by completing the square is only possible in equations containing three parameters.

For example, with an equation like:

$X^2+5X=0$
it will not be possible to solve by completing the square.

Although in most cases we will use the quadratic formula, we recommend getting familiar this method, as it could save you time.

But let's take a step back. What is an equation?

Equations form the base of most mathematics today, and there are many different types of them. They can be as simple as basic addition equations and as complex as the complicated equations used in advanced physics.

Many of the equations we will encounter in our studies will have variables, which are unknown values, usually represented by the letter $X$.

Linear equations, also called first degree equations, have a variable to the power of one. There are also second degree equations, like the quadratic equation, where the variable is squared (to the power of two).

With linear equations (first degree equations), the simplest method to solve them is by isolating the unknown variable.

To do this we must first get rid of any parentheses.

Then, we can move the unknown variable to one side and the known numbers to the other sides using inverse operations.

Then, we reduce to like terms and find the value of the unknown variable.

In order to make sure we arrived at the correct answer, we take our solution and substitute it into the original equation. If both sides of the equation are equal, our solution is correct.

Before diving deeper into quadratic equations, let's review some important terms:

• Expression: an expression is another name for an equation, a formula, or basically any mathematical exercise.
• Factor: a factor is any type of number or variable in an expression/equation.
• Coefficient: a coefficient is a number or a constant that comes before a variable that multiplies the variable in an expression.
  For example, in the expression $4X+5Y+2$
  $4$ is the coefficient of $X$
  and $5$ is the coefficient of $Y$
  Note that the coefficient will not always be a number. Sometimes coefficients also appear as a parameter to be found. For example $aX+bX$. In this case a and $b$, are the coefficients of $X$.
  
• Variable: as previously mentioned, a variable is an unknown value that is usually represented by the letter $X$, but can also be represented by any other letter, as in equations with two unknowns. For example: $X+Y=3$
• Parameter: a parameter is similar to a variable. It is a quantity whose value is unknown but is used in an expression as a known value in order to solve the equation. It is usually represented by letters. However, unlike a variable, its value does not change.
• Side of an equation: in an equation there are two sides seperated by the equal sign. Everything on the left side of the equation is called the left side of the equation, and everything on the right side is called the right side of the equation. Often, we will be required to perform the same operation on both sides in order to solve or simplify an equation.

Let's explain what a Linear, or first degree, equation is:

A linear equation, or a first degree equation, is a way of describing a straight line and its position relative to a system of axes, using mathematical language.

(It is called a first degree equation because the highest exponent is one - we will go more into depth on this later.)

These axes consist of a horizontal and vertical plane that intersect each other and form four squares that divide the system into a positive axis ($+$) and a negative axis ($-$). For example, the equation $Y=X+2=0$ describes what is happening directly on the axes. When $Y$ is equal to zero, $X=-2$. This means that at the point $0$ on the $Y$ axis, the line intersects the $X$ axis
at the point $-2$.

Let's take a look at the following example:

If we have the expression: $X-3=6$
$X-3$ is the left side

• and $6$ is the right side.
  
  Suppose we are asked to find the value of the variable $X$ in the above equation. To do this, we can add $3$ in both members to find out what our variable is. In this case, the expression would look like this:
• $X-3+3=6+3$

So the answer is

• $X=9$
  
• Reduction: when we want to rewrite an expression into a simpler form, we do something called reduction. Suppose we have the expression $X+X+X$ We can add the variables together and write them as $3X$.

We have the equation
$X+5Y+3X+2Y$

After reducing the terms, it will look like this: $4X+7Y$

• Exponents: the exponent of a number tells us how many times the number is multiplied by itself. To express this we use the phrase 'to the power of.' For example: 3 to the power of $2$ (also called $3$ squared) is actually $3$ times $3$. So 3 squared equals $9$.
• Square root: the square root is the number that, mutiplied by itself, gives us the square number. To find the square root of a number, we need to find out what number, multiplied by itself, gives us our square number. For example, $√16$ is equal to $4$. Since $4$ times $4$ (or $4$ to the power of $2$) equals $16$. Note that the numbers obtained by a square root are not always whole numbers. For example, $√10$ will give us $3.16227766017$.
• Parabola: A parabola is a type function with a U-shaped curve. The graph of a quadratic, or second degree, equation takes the shape of a parabola.

Now that we have gone over some important terms, let's see what happens when we use exponents in our equations.

Let's take a look at the following equation. How would you solve it?

$X^2+10X-4=0$

Our intuition tells us that to get rid of the squared X we should solve for the square root. However, if we choose to do that we will have to find the square root of everything in the equation, which would give us decimal fractions and make our equation more complicated than when we started.

The quadratic formula (and other solutions), allow us to solve quadratic equations (second degree equations) in a relatively simple and efficient way. Here we will learn the two most common methods to solve this type of equation.

The square root operation reverses the exponent operation. Finding the square root means to divide a factor by itself. For example, the square root of $X^2$ is simply $X$.

In order to solve the equation without putting the whole equation under the square root and getting lost in a series of complicated calculations, we will use the quadratic formula.

The quadratic equation contains three parameters:

• Parameter $a$ represents the position of the vertex of the parabola on the $Y$ axis. A parabola can have a maximum vertex, or a minimum vertex (depending on if the parabola opens upwards or downwards).
• Parameter $b$ represents the position of the vertex of the parabola on the $X$ axis.
• Parameter $c$ represents the point of intersection of the parabola with the $Y$ axis.

These three parameters are expressed in quadratic equations as follows:

$aX^2+bX+c=0$

A quadratic (second degree) equation tells us a completely different story than the linear equation (first degree equation). Instead of a straight line (like in linear equations), here it is a parabola. A parabola is a function, drawn as a U-shaped curve, that intersects with the $X$ and $Y$ axes at different points. A parabola will either open upwards or downwards. If the equation has a solution, the values of $X$ found are the points of intersection of the parabola with the $X$ axis. For example, if we have the following quadratic equation:

$3X²+8X+4=0$

$X²+2X+0=0$

The equation above gives us our coefficients.

To use the quadratic formula, all we need to do is to plug the parameters into the formula, using the coefficients from the equation.

$X²+2X+0=0$
The values of the parameters will be:
$a =1$
$b =2$
$c =0$

Now we only need to formulate it like this:

Now it will be much simpler.

Therefore $x_1=0$

To find the $x_2$ , in our formula we have to subtract as follows:

Therefore $x_2=-2$

If we perform the operations step by step, working carefully, we will arrive at a precise result.

No solution?

Did you get a negative number in the square root?

Maybe the parabola doesn't intersect the X axis.

But, to be safe -> check again. It is important to proceed carefully to get an accurate result. Graphing calculators can be useful to check that we are in the right direction.

Your solution should be: $X=0$ y $X=-2$
which means that the intersection points on the axis $X$ will be at $-2$ and $0$.

If so, a quadratic equation can give us:

• two solutions (two intersections of the function with the $X$ axis)

Other examples of graphs of quadratic functions

When there is a solution
Parabola tangent to the axis $X$, as in the equation:
$Y=X^2+2X+1$

No solution
Parabola "floats" on the axis $X$, as in eq:
$2=X^2+2X+2$

How to solve quadratic equations without a calculator?

In most cases, we will use the quadratic formula.

The quadratic formula is the fastest and most efficient way to solve quadratic equations.

Why do we call quadratic equations second degree equations?
In a quadratic equation we apply an exponent to our variable $X^2$. This is also called a second degree variable because the highest exponent in quadratic equations will be two. Let's take a closer look at exponents.

As we have already mentioned, an exponential number is a number multiplied by itself. Instead of writing $X\times X$, we write it like this: $X^2$. The larger number on the left is called the base. The small number to the right of the base, called the exponent, tells us how many times the base must be multiplied by itself.

Although this method will not work in every case, if applicable, it can provide us with a quick solution to the equation we need to solve.

It is important to note that, unlike the quadratic formula, solving an equation with the perfect square method is only possible in equations that contain three coefficients.

For example, in the case of $X^2+5X=0$ it will not be possible to solvethe equation using the perfect square method. We recommend being familiar with this method, although in most cases we will use the quadratic formula.

Sometimes, we will not be asked for the specific points where the parabola intersects the $X$ axis, but only how many points of intersection there are.

By calculating the value of the discriminant, we will know how many solutions to look for.

The discriminant, represented by the symbol delta, represents change. In the quadratic formula the discriminant is what appears just below the square root symbol.

If $> 0$ positive => the equation will have two solutions.

If $= 0$ the equation will have one solution.

If $< 0$ negative => the equation will have no solution.

Do you want to test yourself? Or maybe you already understand the material and are trying to solve exercises quickly?

Try using a calculator to solve quadratic equations. All you have to know is how to input the missing values in the equation, and you will get the solution.

Task:

Solve the following equation:

$X^2+5X+4=0$

Solution:

$a= 1$

$b= 5$

$c= 4$

We input the values of $a,b,c$ of the given equation into the quadratic formula:

$X_{1,2\text{ }}=\frac{-b±\sqrt{b²-4\cdot a\cdot c}}{2\cdot a}$

$X_{1,2\text{ }}=\frac{-5±\sqrt{5²-4\cdot1\cdot4}}{2\cdot a}$

$X_{1,2\text{ }}=\frac{-5±\sqrt{25-16}}{2}$

$X_{1,2\text{ }}=\frac{-5±\sqrt{9}}{2}=\frac{-5±3}{2}$

Answer:

The correct answer is:

$X_1=-1,\text{ X}_2=-4$

Task:

Solve the following equation:

$2X²-10X-12=0$

Solution:

$a=2$

$b=-10$

$c=-12$

We input the values of $a,b,c$ of the given equation into the quadratic formula:

$X_{1,2}=\frac{-b±\sqrt{b²-4\cdot a\cdot c}}{2\cdot a}$

$X_{1,2}=\frac{-(-10)±\sqrt{(-10)²-4\cdot2\cdot(-12)}}{2\cdot2}$

$X_{1,2}=\frac{10±\sqrt{100+96}}{4}$

$X_{1,2}=\frac{10±\sqrt{196}}{4}=\frac{10±14}{4}$

$X_1=\frac{10+14}{4}=\frac{24}{4}=6$

$X_2=\frac{10-14}{4}=\frac{-4}{4}=-1$

Answer:

The correct answer is:

$X_1=6,\text{ X}_2=-1$

Task:

Solve the following equation:

$-81X^2+54X-9=0$

Solution:

$a=-81$

$b=54$

$c=-9$

We input the values of $a,b,c$ of the given equation into the quadratic formula:

$X_{1,2}=\frac{-b±\sqrt{b²-4\times a\times c}}{2\times a}$

$X_{1,2}=\frac{-54±\sqrt{54²-4\times (-81)\times(-9)}}{2\times(-81)}$

$X_{1,2}=\frac{-54±\sqrt{2916-2916}}{-162}$

$X_{1,2}=\frac{-54±\sqrt{0}}{-162}=\frac{-54}{-162}=\frac{1}{3}$

Answer:

$X=\frac{1}{3}$

Task:

Solve the following equation:

$-6X²+12X-14X=0$

Solution:

The number under the square root symbol is a negative number. Since a negative number has no square root, this equation has no solution.

Answer:

There is no solution.

Task:

Given triangle $ABC$:

Given:

$a+b=7$

We are given that the ratio between $CB$ and $AC$ is $5:3$

Task:

Find the length of $a$ and $b$.

Solution:

Use the Pythagorean theorem

$AB²+AC²=BC²$

for $\triangle ABC$

Because:

$\frac{CB}{AC}=\frac{5}{3}$

We know that:

$\frac{CB}{a}=\frac{5}{3}$

Therefore, to reduce we must multiply both terms $\times a$

Which gives us:

$CB=\frac{5}{3}a$

Now we can input the values into the quadratic equation:

$b²+a²=(\frac{5}{3}a)²$

$a+b=7$,$b=7-a$

$(7-a)²+a²=(\frac{5}{3}a)²$

$7²+a²-2\times7\times a+a²=\frac{25}{9}a²$

$2a²-14a+49=\frac{25}{9}a²$

$0=\frac{7}{9}a²+14a-49$

Using the quadratic formula, we get:

$X_{1,2}=\frac{-b±\sqrt{b²-4\times a\times c}}{2\times a}$

$X_{1,2}=\frac{-14±\sqrt{14²-4\times\frac{7}{9}\times(-49)}}{2a}$

$X_{1,2}=\frac{-14±\sqrt{\frac{3136}{9}}}{\frac{14}{9}}=-\frac{14±\frac{56}{3}}{\frac{14}{9}}$

$X_1=(-14+\frac{56}{3}):\frac{14}{9}=\frac{14}{3}\times\frac{9}{14}=\frac{9}{3}=3$

$a=3\operatorname{cm}$

$b=7-3=4\operatorname{cm}$

$X_2=(-14-\frac{56}{3}):\frac{14}{9}=\frac{98}{3}\times\frac{9}{14}=-21$

Remember that the answer cannot be negative since it represents the length of a side.

Answer:

$a=3\operatorname{cm}$

$b=7-3=4\operatorname{cm}$

Task:

Solve the following equation:

$10x^2-65x+135=4x^2+13x-45$

Solution:

First, we have to rewrite the equation in standard form

$aX²+bX+C=0$

By moving all of the numbers and variables to the left side:

$10X²-65X+135=4X²+13X-45$ / Shift

$-4X²-13X+45$

$10X²-4X²-65X-13X+135+45=0$

$6X²-78X+180=0$

Our parameters are:

$a=6$

$b=-78$

$c=180$

We input the parameters into the quadratic formula:

$X_{1,2}=\frac{(-78)±\sqrt{ \left(-78\right)^2 -4\cdot6\cdot180}}{2\cdot6}$

Answer:

$X_1=-10$

$X_2=-3$