Imaginary Roots For Quadratic Equation

Solving Quadratic Equations

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 thenumber 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 begiven as an answer to obtain full marks.

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