Math 110
Chapter 1, Section 3
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Equation Review: Quadratic Equations (§1.3)
ax 2  bx  c  0
(You may need to rearrange things in order to get things looking like the above)
Discriminant:
b 2  4ac
 If this is > 0, then there are two real answers
 If this is == 0, then there is one real answer
 If this is <= , then there are no real answers (but may be complex answers)
Solving By Factoring:
2
1. Take the given, 2nd degree polynomial equation: x  x  20  0
2. Factor into two, 1st degree polynomials x  5x  4  0
3. Set each polynomial to be independently equal to zero:
x  5  0
x  4  0
4. Solve each (new) equation independently:
x = -5
x=4
Notice that after step #3, we've reduced the original, challenging problem (solving a
quadratic equation) to a easier, previously solved problem (solving a linear equation(s) ).
Solving By Completing The Square:
Before explaining the 'completing the square' bit, first explain the idea of a perfect square.
2
b
Have them factor a couple of perfect squares, so that the pattern is clear, and it's obvious where the   term comes from.
2
Perfect Squares:
x  a 2  x 2  2ax  a 2 x  a 2  x 2  2ax  a 2
1. Given the starting, quadratic equation, you first need to eliminate the leading coefficient
(if any).
So if a isn't 1, then divide everything by a)
2. Move c over to the right-hand-side (RHS)
2
b
3. Add   to both sides. If b is negative, then this term needs to be negative, too.
2
Since the next step is to factor the LHS, you can leave the thing you're adding in the
2
2
b
 b 
format of   (Technically, if you already divided by a, you'll have   )
2
 2a 
2
4.
5.
6.
7.
b

Factor the Left H.S. – You'll get  x  
2

At this point, you're almost done – sum up the RHS,
Take the square root of both sides (don't forget the ± !!!)
Solve for the +, and for the -, and you're done!
Math 110
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Math 110
Chapter 1, Section 3
Page 2 / 2
Solving By The Quadratic Formula:
1. Given the starting, quadratic equation, you first need to get it into the 'standard form'
listed at the top.
2. Next, plug-and-chug with this formula:
X
 b  b 2  4ac
x
2a
10
9
8
7
6
5
4
3
2
1
0
-1 0 1
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-2
0
-3
-4
-5
-6
-7
-8
-9
-10
2 3 4 5 6 7 8 9 10
Y
Math 110
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