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Welcome to class!
In today’s class, we will be talking about quadratic equation by factorization and completing the square method. Enjoy the class!
Quadratic Equation by Factorization and Completing the Square Method
Solving quadratic equations by factoring
The general form of a quadratic equation is
ax2 + bx + c = 0
where x is the variable and a, b & c are constants.
Examples of quadratic equations
(a) 5x2 − 3x − 1 = 0 is a quadratic equation in quadratic form where a=5, b=−3, c=−1
(b) 5 + 3t − 4.9t2 = 0 is a quadratic equation in quadratic form.
Here, a=−4.9, b=3, c=5
[This equation arose from finding the time when a projectile, being acted on by gravity, hits the ground.]
(c) (x + 1)2 = 4 is a quadratic equation but not in quadratic form.
It has to be expanded and simplified to:
x2 + 2x − 3 = 0
Summary
In general, a quadratic equation:
- must contain an x2 term
- must NOT contain terms with degrees higher than x2 e.g. x3, x4 etc.
Examples of non-quadratic equations
- bx − 6 = 0 is NOT a quadratic equation because there is no x2 term.
- x3 − x2 − 5 = 0 is NOT a quadratic equation because there is an x3 term (not allowed in quadratic equations).
Solutions of a quadratic equation
The solution of an equation consists of all numbers (roots) which make the equation true.
All quadratic equations have 2 solutions (i.e. 2 roots). They can be:
- real and distinct
- real and equal
- imaginary (complex)
Example 1:
The quadratic equation x2 − 7x + 10 = 0 has roots of x=2 and x=5. (We’ll show below how to find these roots.)
This can be seen by substituting in the equation:
When x = 2,
x2 − 7x + 10
= (2)2 − 7(2) + 10
= 4 − 14 + 10
= 0
(This can be shown similarly for x = 5). In this example, the roots are real and distinct.
Example 2:
The quadratic equation x2 − 6 x + 9 = 0 has double roots of x = 3 (both roots are the same)
This can be seen by substituting x = 3 in the equation:
x2 − 6x + 9
= (3)2 − 6(3) + 9
= 9 − 18 + 9
= 0
Example 3:
The quadratic equation
x2 + 9 = 0
has imaginary roots of x = √−9 or − √−9
Example 4:
Solve x2 − 2x − 15 = 0
x2 − 2x − 15 = 0
Factoring gives
(x − 5) (x + 3) = 0
Now, if either of the terms (x − 5) or (x + 3) is 0, the product is zero. So, we conclude:
(x − 5) = 0, therefore
x = 5
or
(x + 3) = 0, therefore
x = − 3
Hence the roots are x = 5 and x = − 3.
Are we correct?
We check the roots in the original equation by substitution.
When x = 5:
x2 − 2x − 15
= (5)2 − 10 − 15
= 25 − 10 − 15
= 0
(Similarly, when we substitute x = −3, we also get 0.)
Example 5:
Solve
9x²+6x+1=0
9x2 + 6x + 1 = 0
Factoring gives
(3x + 1) (3x + 1) = 0
So, we conclude:
(3x + 1) = 0,
Therefore x = −1/3
We say there is a double root of x = −1/3
In our next class, we will be talking about the General Form of Quadratic Equation Leading to Formula Method. We hope you enjoyed the class.
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
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