- algebra help with quadratic equations
- grab the basics solving quadratic equarions
- additional mathematic quadratic function
- explain how to check quadratic equations
- online simutaneous equation solver
- rules in solving quadratic equations
- written explanation on how to solve quadratic formulas
- tell me something about about quadratic equation
- simultaneous equations 3 unknowns
- what are the pros and cons of quadratic formula, completing the square and factoring
- explain factorise equations
- "jericho demo walkthrough"
- free worksheet on solving qudractic equations
- quadratic equation solver solve for y
- solving quadratic simultaneously equations
- fx-3650p factorization programme
- similtaneous equation
- make quadratic formula
- what is with quadratic equations
- equation calculator with workings
|
|
Additional Maths Quadratic 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
