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 $\displaystyle {{x}_{1}}$
$\displaystyle {{x}_{2}}=\frac{-\left( -3 \right)+\sqrt{1}}{2\cdot 2}$
$\displaystyle {{x}_{2}}=\frac{3+1}{4}$
$\displaystyle {{x}_{2}}=\frac{4}{4}$
x2 = 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 = 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.
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
$\displaystyle {{\sqrt{\left( x+1 \right)}}^{2}}=\pm \sqrt{6}$
$\displaystyle \left( x+1 \right)=\pm \sqrt{6}$
x1 = $ \displaystyle \sqrt{6}-1$ or x1 = $ \displaystyle -\sqrt{6}-1$