Back To The Basics Of Quadratic Equations
May 9th, 2008

Looking at an example:

You are given the question

x²+5x+4=0

In the above example,

a=1 (If there is no number before the x then we can assume that the number is 1)

b=5

c=4

What we then have to do is find out what x could be equal to in order to satisfy the equation and therefore make it true. In this example, x can either equal -4 or -1 . We don't know yet how we came to this answer, but let's show that both of these answers do work.

Putting -4 into the above equation:

x²+5x+4=0

(-4)² + (5*-4) + 4 = 0

(-4)*(-4)  + (5*-4) + 4 = 0

16 + (-20) + 4 = 0

16-20+4=0

-4+4=0

0=0

This shows that -4 can be a solution for x

Putting -1 into the above equation:

x²+5x+4=0

(-1)² + (5*-1) + 4 = 0

(-1)*(-1)  + (5*-1) + 4 = 0

1 + (-5) + 4 = 0

1-5+4=0

-4+4=0

0=0

This shows that -1 can also be a solution for x

Therefore, there are two possible solutions, -1 and -4. They must both be given as an answer to obtain full marks.

Once you have obtained the possible solutions for x, is is always necessary to check them in the above way

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