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**Quadratic equation**

**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, **a**, **b**, **c** 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

**Method 1:**

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:

** **

**The second case:**

If D = 0 then we have one real root.

D = b² – 4ac = 0

Therefore, the formula consists on:

**The third case:**

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

**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.**

D = 9 – 8

**D = 1**

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

**Find the roots:**

**So, we found **

x_{2} = 1

**We found **our 2 different real roots:

and

**b) **2x² – 4x + 2 = 0

D = 16 – 16

D = 0

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

**x = 1**

So, we found the root x = 1.

**c) **2x² – 3x + 3 = 0

D = b² – 4ac

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

Then, solve each of those equations.

(x – 6) = 0

**x = 6**

(x + 6) = 0

**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 ?

**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

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 4 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)

32 = (-2) x (-16)

32 = 4 x 8

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.**

or

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 3****: **Complete 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

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

x_{1} = or x_{1} =