Quadratic equations

A quadratic equation is an equation that contains one unknown that is multiplied to the other. If an equation contains an unknown that is multiplied to a higher exponent than the other, then it is no longer a quadratic equation.

Description of a quadratic equation

The basic form of a quadratic equation is as follows:

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

The values of a, b, c are real numbers and the value of a is different from zero. Other names:

• ax² is called the quadratic term,
• bx is called the linear term,
• c is called the absolute term.

Any quadratic equation can be modified to its basic form by using equivalent modifications or other modifications. An example of a quadratic equation might be the following equation:

$$3x^2-6x+8=0.$$

A more difficult question may already be whether this is also a quadratic equation:

$$(x+1)\cdot(x+2)=7.$$

You can't see x² anywhere, and there is no zero on the right-hand side. To solve the right-hand side, simply subtract seven to get the form

$$(x+1)\cdot(x+2)-7=0.$$

There is still no x². We multiply the parentheses and get:

$$x^2+2x+x+2-7=0$$

We now have x². Adding them up, we finally get the basic form of the quadratic equation:

$$x^2+3x-5=0.$$

What about the following example, is it a quadratic equation? Can it be converted to its basic form?

$$x^2+3x+\frac{4}{x}=0$$

The answer is that it can't. Because if we multiply the whole equation by the variable x, to get rid of the variable in the denominator of the fraction, we get:

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

Now we have x³, which cannot be in a quadratic equation.

There are special types of quadratic equations, depending on what value the coefficients a, b, and c.

The values of the coefficients

For the next chapters, it is absolutely critical that you can identify the coefficients of a, b, and c. Let's start with a simple example. Determine the coefficients for the equation 3x² + 4x + 7 = 0. Here's an easy one, a = 3, b = 4, and c = 7.

What about, for example, the equation 3x² − x + 2 = 0? The values of a and c are clear, a = 3 and c = 2, but what about b? The mistake would be to write that b = 1. Imagine that x is preceded by a one, like this: 3x² − 1x + 2 = 0. Now it might be more obvious that b = −1.

Another interesting example: x² − 3 = 0. Here again there may be a problem with the coefficient of b, the others are obvious: a = 1 and c = −3. What is the value of the coefficient of b? The linear term does not appear in this equation at all, it is the same as if we wrote the equation in the form x² + 0x − 3 = 0. So it is true that b = 0.

A more complicated example:

$$5x^2+3x+\pi\cdot x=0$$

There are two betrayals in this equation. There is no absolute term, so c = 0. The quadratic term is obvious, so a = 5. But what about the cursed b? We can tweak the equation some more, we'll extract the variable x from the two expressions as follows:

$$5x^2+(3+\pi)x=0$$

The coefficient b is again what is "before" the variable x, so the whole parenthesis is equal to the coefficient b = (3+π). Similar equation:

$$-x^2+4x+1-\sqrt{2}=0$$

The first two coefficients are obvious: a = −1 and b = 4. What is the absolute term? These are the remaining terms where there is no x. So, the $c=1-\sqrt{2}$. Don't worry about the square root, it doesn't bite, it's a normal number.

And what are the values for this equation? (x + 1)(x + 2) = −2x. First we have to modify the equation to its basic form. We multiply the parentheses and we get

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

and finally add to the equation 2x to get the zero on the right hand side:

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

The values are already obvious: a = 1, b = 5, c = 2.

A purely quadratic equation

If b = 0 is a purely quadratic equation, it is solved similarly to a linear equation. This equation therefore has the basic form:

$$ax^2+c=0$$

Since b = 0, the entire linear term bx drops out of the equation because 0x = 0. We solve the equation by first subtracting the absolute term. That is, we subtract the value of c from the equation. This gives us:

$$\begin{eqnarray} ax^2+c&=&0\quad/-c\\ ax^2&=&-c \end{eqnarray}$$

Next, we divide the equation by a. We get the form:

$$x^2=\frac{-c}{a}$$

Finally, we subtract the whole equation. The fraction on the right hand side must not be negative, because you cannot subtract a negative number. We get the result:

$$x_1=\sqrt{\frac{-c}{a}},\quad x_2=-\sqrt{\frac{-c}{a}}$$

Note that we get two results, one positive and one negative. This will be better seen in the example. Let's calculate the equation

$$2x^2-8=0.$$

Following the procedure, we first subtract the member c. Note that the member c is possibly negative in this case, so we are subtracting a negative number, thus adding a positive one. Thus, we add +8 to the whole equation (the goal is to get the 8 on the right side). We get the form:

$$2x^2=8$$

Next, we follow the procedure to divide the equation by a, which is two. After dividing, we get the form:

$$x^2=4$$

Now subtract.

$$x_1=2,\quad x_2=-2$$

Why is there a minus two? Because minus two times minus two is four. We can also look at the graphical solution: we can see that the graph of the function 2x² − 8 = 0 intersects the axis x just at the points x₁ = −2 and x₂ = 2.

Solve the equation

$$80x^2-5=0.$$

Add five to the equation:

$$80x^2=5$$

Divide by 80:

$$x^2=\frac{5}{80}$$

Truncate the fraction:

$$x^2=\frac{1}{16}$$

Subtract:

$$x_1=\frac14,\quad x_2=-\frac14$$

Quadratic equation with no absolute term

Another special case of a quadratic equation occurs when the absolute term c is zero. The equation has the basic form:

$$ax^2+bx=0$$

This type of equation is solved by displacing the unknown:

$$x(ax + b) = 0.$$

Here we can already see a fairly clear result. Again, two roots of the equation will come out, but one of them will always be equal to zero. In fact, the whole left-hand side is in the form of the product of two expressions: x and then the parenthesis (ax + b). When is this product equal to zero? When at least one of these expressions is equal to zero. So the left-hand side is zero in two cases:

$$x=0\quad\mbox{ Or }\quad ax+b=0.$$

We already have the first result of the equation, so it is zero. We get the next one if we solve the linear equation ax + b = 0. We know from the article on linear equations that this equation has exactly one solution that is equal to:

$$x=-\frac{b}{a}$$

So we can now write down the result of the whole quadratic equation ax² + bx = 0:

$$x_1=0,\quad x_2=-\frac{b}{a}.$$

Example:

$$6x^2+3x=0$$

Let's print x:

$$x(6x+3)=0$$

The first solution is x = 0. We find the second solution by solving the linear equation 6x + 3 = 0:

$$x=-\frac{b}{a}=-\frac36=-\frac12.$$

Let's write down the result:

$$x_1=0,\quad x_2=-\frac12.$$

Again, we can look at the graph of the function that intersects the x axis at these points:

Other ways of solving

These were the basic techniques for specific types of quadratic equations. A general quadratic equation is solved using a discriminant. Other types of equations include a quadratic equation with a parameter.