Quadratics And Simultaneous Equations

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

GCSE Module 5

(Click the thumbnails to make bigger) Quadratics Graphs Co-ordinate Geometry Trigonometry Simultaneous Equations Manipulating Expressions and Equations oh here is a good site to practise those BASICS, which seem to be letting everyone down, it's a cool games site. You'll like it trust me and just because everyone loves him, here's math dragon Right Here's my first video attempt, it's a bit rubbish and no sound but it will get better!