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## What is a quadratic equation?

A quadratic equation is an equation of degree 2 with one variable and that can be written as:
x² + bx + c = 0, where a≠0, abc are the coefficients and x is the variable.
The number of roots of a quadratic equation is less or equal to the number of its degree.
The graph of a quadratic equation is shaped in the U form and it’s called a Parabola.

### How to solve a quadratic equation?

Three methods you can use for solving a quadratic equation:

Method 1: Quadratic formula

Method 2: Factoring

Method 3: Completing the square

#### Method 1: Quadratic formula

The first method is the more used and easy to apply if it’s easy for you to memorize formulas.

When solving a quadratic equation we should be aware of three cases.
Those three cases are distinguished by a formula called the “Discriminant”.

The discriminant is the value under the radical sign:
D = b² – 4ac

The first case:

If  D > 0,  then we have two different real roots.

D = b² – 4ac > 0

Therefore, the formula consists on:

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

The second case:

If D = 0 then we have one real root.

D = b² – 4ac = 0

Therefore, the formula consists on:

$\displaystyle x=-\frac{b}{2a}$

The third case:

If D < 0, then we don’t have real roots, we say that the roots are imaginary.

$\displaystyle D={{b}^{2}}-4ac<0$

Worked Example 1: Solve the Equations
a) 2x² -3x + 1 = 0

b) 2x² – 4x + 2 = 0

c) 2x² – 3x + 3 = 0

## Solution:

a) 2x² -3x + 1 = 0
Determine the coefficients first.

a = 2

b = -3

c = 1

## Find the discriminant.

$\displaystyle D={{\left( -3 \right)}^{2}}-4\cdot 2\cdot 1$
D = 9 – 8
D = 1

Since the D > 0, then we already know that we have 2 solutions, 2 real roots.

Find the roots:
$\displaystyle {{x}_{1}}=\frac{-\left( -3 \right)-\sqrt{1}}{2\cdot 2}$

$\displaystyle {{x}_{1}}=\frac{3-1}{4}$

$\displaystyle {{x}_{1}}=\frac{2}{4}=\frac{1}{2}$

So, we found $\displaystyle {{x}_{1}}$

$\displaystyle {{x}_{2}}=\frac{-\left( -3 \right)+\sqrt{1}}{2\cdot 2}$

$\displaystyle {{x}_{2}}=\frac{3+1}{4}$

$\displaystyle {{x}_{2}}=\frac{4}{4}$

x2 = 1

We found our 2 different real roots:

$\displaystyle {{x}_{1}}=\frac{1}{2}$ and $\displaystyle {{x}_{2}}=1$

b) 2x² – 4x + 2 = 0

$\displaystyle D={{4}^{2}}-4\cdot 2\cdot 2$

D = 16 – 16

D = 0

Since D = 0, then we have only one real root.

$\displaystyle x=-\frac{b}{2a}$

$\displaystyle x=-\frac{(-4)}{2\cdot 2}$
x = 1
So, we found the root x = 1.

c) 2x² – 3x + 3 = 0

D = b² – 4ac

$\displaystyle D={{3}^{2}}-4\cdot 2\cdot 5$

D = 9 – 30

D = -21

Since D < 0 then we don’t have real roots.

#### Method 2: Factoring

The steps of solving a quadratic equation by factoring:

Step 1: First put all the terms on the left side on descending order and make the equation equal to zero.

Step 2: Factor by using one of the factoring strategies.

Step 3: Make each factor equal to zero.

Step 4: After that, solve each of those equations.

Step 5: Then to be sure we have found the right roots we try if our roots make the first equation true.

Worked example 2: Solve x² – 36 = 0 Solution: This is a quadratic equation when the b coefficient is missing. Our equation is on the right form. x² – 36 = 0

Using the method ‘’The difference of two perfect squares’’ a² – b² = (a-b)(a+b)

x² – 36 = 0

(x – 6)(x + 6) = 0

Make each factor equal to zero.

(x – 6) = 0 or (x + 6) = 0

x = 6 or x = -6

##### The proof.

To check if x = 6 or x = -6  we try if they make our equation true:

(-6)² – 36 = 0

36 – 36 = 0

0 = 0 or (6)² – 36 = 0

36 – 36 = 0

0 = 0

So,  6 is the root of the equation.

Worked Example 3: Solve $\displaystyle {{x}^{2}}+13x=3x$? Solution: First put all the terms on the left side on descending order and make the equation equal to zero.

x² + 13x = 3x

x² + 13x – 3x = 0

$\displaystyle x{}^\text{2}\text{ }+\text{ }10x\text{ }=\text{ }0$.

Now all you have to do is factor the common variable:

x² + 10x = 0

x(x + 10) = 0

Make each factor equal to zero and then solve each of this equations.

x = 0 or x + 10 = 0

x = 0 or x = -10

##### The proof

0² + 10 ⋅ 0 = 0 or (-10)² + 10⋅(-10) = 0

0 = 0 or 100 – 100 = 0

0 = 0 or 0 = 0

So numbers 0 and -10 are the roots.

Worked Example 4: Solve x² – 12x + 32 = 0

Solution: Our equation is on the right form.

x² – 12x + 32 = 0

Using the method:

Step 1: Determine the coefficients

a = 1;  b = -12;  c = 32;

Step 2: Find two numbers that multiply gives ac and add to give b.

To find this numbers you have to write c as all the possibilities of a product of two integers.

32 = 1  x  32

32 = (-1)  x  (-32)

$\displaystyle 32\text{ }=\text{ }2~\text{ }x~\text{ }16$

32 = (-2)  x  (-16)

32 = 4  x  8

$\displaystyle 32=\left( -4 \right)x~\left( -8 \right)$

Now, from all this products find those two numbers that multiply gives ac and add give b.

We see that the numbers we are looking for are (-4) and (-8), because 32 = (-4) x (-8) and when we add (-4) + (-8) = -12 .

Then we write our equation as a product of two first grades equation with the two numbers we found.

x² – 12x + 32 = (x – 4)(x – 8)

Make each factor equal to zero.

$\displaystyle \left( x-4 \right)=0$  or $\displaystyle \left( x-8 \right)=0$

Then, solve each of those equations.

x – 8 = 0

x – 4 = 0

x= 8 or x = 4

The proof:

x² – 12x + 32 = 0

(4)² – 12⋅4 + 32 =0

16 – 48 + 32 = 0

-32 + 32 = 0

(8)² – 12⋅8 + 32 =0

64 – 96 + 32 = 0

-32 + 32 = 0

0 = 0

So, we say that x = 4 and x = 8 are the roots.

#### Method 3: Completing the square

Step 1: Divide all the terms with (a) the coefficient of  x²

Step 2: Then move the term (c/a) to the right side of the equation.

Step 3Complete the square on the left side by adding the value you need and don’t forget to add it to the right side to obtain the equality.

Step 4: Take the square root on the both sides and find the values of x.

##### Worked Example 5: Solve 3x² + 6x – 15 = 0

Solution: Divide all the terms with the ‘a’ coefficient:

a = 3, when we divide with 3 we obtain:

3x² + 6x – 15 = 0

$\displaystyle \frac{3{{x}^{2}}}{3}+\frac{6x}{3}-\frac{15}{3}=0$

x² + 2x – 5 = 0

Move the term (c/a) to the right side of the equation.

x² + 2x = 5

##### Complete the square.

x² + 2x = 5

To complete the square on the left side we need to add the number 1 and we do that to the right side two to obtain the equality.

x² + 2x + 1 = 5 + 1

(x + 1)² = 5 + 1

(x + 1)² = 6

##### Take the square root on the both sides and find the values of x.

(x + 1)² = 6

$\displaystyle {{\sqrt{\left( x+1 \right)}}^{2}}=\pm \sqrt{6}$

$\displaystyle \left( x+1 \right)=\pm \sqrt{6}$

x1 = $\displaystyle \sqrt{6}-1$ or  x1 = $\displaystyle -\sqrt{6}-1$