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Solving Quadratic Equations


A quadratic equation is an equation where the exponent of the variable is at most 2.

The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and solutions. You can’t conclude anything about the individual terms of the unfactored quadratic (like the 5 x or the 6), because you can add lots of stuff that totals zero.

So the first thing I have to do is factor: x2 + 5x + 6 = (x + 2)(x + 3)

Set this equal to zero: (x + 2)(x + 3) = 0

Solve each factor: x + 2 = 0 or x + 3 = 0 x = –2 or x = – 3

The solution to x2 + 5x + 6 = 0 is x = –3, –2

Checking x = –3 and x = –2 in x2 + 5x + 6 = 0:

[–3]2 + 5[–3] + 6 ?=? 0

9 – 15 + 6 ?=? 0

9 + 6 – 15 ?=? 0

15 – 15 ?=?

0 0 = 0

[–2]2 + 5[–2] + 6 ?=? 0

4 – 10 + 6 ?=? 0

4 + 6 – 10 ?=? 0

10 – 10 ?=? 0

0 = 0

Exercise 2:

Quadratic Equations This equation is not in “(quadratic) equals (zero)” form, so I can’t try to solve it yet. The first thing I need to do is get all the terms over on one side, with zero on the other side. Only then can I factor and solve:

x2 – 3 = 2x

x2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x – 3 = 0 or x + 1 = 0

x = 3 or x = –1


Then the solution to x2 – 3 = 2x is x = –1, 3

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