##### 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

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.

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

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 **$\displaystyle {{x}_{1}}$

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

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

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

x_{2} = 1

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

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

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

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 = 1So, 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.

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.

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.

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 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

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

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

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

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

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