Quadratic Equation
A Quadratic Equation in Standard Form
(a, b, and c can have any value, except that a can't be 0.)
- The letters a, b and c are coefficients (you know those values)
- The letter "x" is the variable or unknown (you don't know it yet)
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Here is an example:
The name Quadratic comes from "quad" meaning square, because of x2 (in other words x squared).
It can also be called an equation of degree 2
More Examples of Quadratic Equations:
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In this one a=2, b=5 and c=3 |
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This one is a little more tricky:
- Where is a? In fact a=1, because we don't usually write "1x2"
- b=-3
- And where is c? Well, c=0, so is not shown.
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Oops! This one is not a quadratic equation, because it is missing x2 (in other words a=0, and that means it can't be quadratic) |
How To Solve It?
Quadratic equations can be solved using a special formula called the Quadratic Formula:
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The ± means you need to do a plus AND a minus, and so there are normally TWO solutions ! |
The answers it gives are the "solutions" to the Quadratic Equation, and are often called "roots", or sometimes "zeros"
Discriminant
The blue part (b2 - 4ac) is called the discriminant, because it can "discriminate" between the possible
types of answer:
- when b2 - 4ac is positive, you will get two solutions
- when it is zero you get just ONE solution, and
- when it is negative you
get two Complex solutions
Remembering
I don't know of an easy way to remember the Quadratic Formula, I just say to myself:
"minus b plus or minus the square root of b-squared minus four ac, all over two a"
Solving
To solve, just plug the values of a, b and c into the Quadratic Formula, and do the calculations.
Example: Solve 5x² + 6x + 1 = 0
| Coefficients are: |
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a = 5, b = 6, c = 1 |
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| Quadratic Formula: |
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x = [ -b ± √(b2-4ac) ] / 2a |
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| Substitute a,b,c: |
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x = [ -6 ± √(62-4×5×1) ] / (2×5) |
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| Solve: |
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x = [ -6 ± √(36-20) ]/10 |
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x = [ -6 ± √(16) ]/10 |
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x = ( -6 ± 4 )/10 |
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x = -0.2 and -1 |
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Answer: x = -0.2 and -1 |
| Check -0.2: |
5×(-0.2)² + 6×(-0.2) + 1
= 5×(0.04) + 6×(-0.2) + 1
= 0.2 -1.2 + 1
= 0 |
| Check -1: |
5×(-1)² + 6×(-1) + 1
= 5×(1) + 6×(-1) + 1
= 5 - 6 + 1
= 0 |
Solve by Factoring
Sometimes it is easy to solve quadratic equation by factoring. Read that page to learn more.
Complex Solutions?
When the Discriminant (the value b2 - 4ac) is negative you get Complex solutions ... what does that mean?
It means your answer will include Imaginary Numbers. Wow!
Example: Solve 5x² + 2x + 1 = 0
| Coefficients are: |
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a = 5, b = 2, c = 1 |
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| The Discriminant is negative: |
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b2 - 4ac = 22 - 4×5×1 = -16 |
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| Use the Quadratic Formula: |
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x = [ -2 ± √(-16) ] / 10 |
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The square root of -16 is 4i, where i is √-1
Read Imaginary Numbers to find out why |
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x = ( -2 ± 4i )/10 |
Answer: x = -0.2 ± 0.4i
In some ways it is actually easier... you don't have to calculate the two solutions, you just leave it as -0.2 ± 0.4i.
Hidden Quadratic Equations!
Sometimes a quadratic equation doesn't look like one.
Your job will be to recognise it and (with a little clever work) turn it into "Standard Form".
Standard Form: ax² + bx + c = 0
Here are some examples for you:
| In disguise |
What to do |
In Standard Form |
a, b and c |
| x2 = 3x -1 |
Move all terms to left hand side |
x2 - 3x + 1 = 0 |
a=1, b=-3, c=1 |
| 2(w2 - 2w) = 5 |
Expand (undo the brackets), and move 5 to left |
2w2 - 4w - 5 = 0 |
a=2, b=-4, c=-5 |
| z(z-1) = 3 |
Expand, and move 3 to left |
z2 - z - 3 = 0 |
a=1, b=-1, c=-3 |
| 5 + 1/x - 1/x2 = 0 |
Multiply by x2 |
5x2 + x - 1 = 0 |
a=5, b=1, c=-1 |
Summary
- Quadratic Equation in Standard Form: ax² + bx + c = 0
- Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a
- When the Discriminant (b2-4ac) is:
- positive, there are 2 solutions
- zero, there is one solution
- negative, there are 2 complex solutions
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