- quad equasion
- lesson plan solving quadratic equations by graphing
- solving quadratic by factorisation
- slavery of quadratic formula
- find formula for quadratic curve
- what are pros and cons to solving quadratic equations
- solving quadratic equation asm
- solving quadratic equations using ti-86
- quadratic curves word problems
- solve simultaneous equation using c program
- math 3a solving quadratic functions
- parabolic sar formula
- step by step solve quadratic equations
- fcat math equation quadratic
- how do i know how to do quadratic equations
- quadratic equations 28real-life examples29
- "equations without any context"
- "x2 - 3x + 2 e289a4 0"
- vba +"solve equations"
- quadratic equation question bank
|
|
Basics Quadratics
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
