- show the calculation for the vertex of the quadratic equation
- solving for c in a quadratic trinomial
- word problem linear equations and quadratic equation
- simultaneous equations solving on line
- solving quadratic equation b
- quadratic equation solution solver
- online simultaneous equation calculator
- quadratic function word problem solving with solution
- quadratic word problems with solutions
- founder of the quadradic formula
- worded problems solving quadratic equation
- quadratic equations squared
- example of solving mdas
- writingquadraticequations
- pros and cons of quadric formula, completing the square, and factoring
- quadratic equation real life application
- importance of fx 991es calculator
- the hardest equation ever
- line of symmetry equation y = -2%5e + 4x + 5
- dhlg-4 pulse
|
|
Solving Quadratics Basics
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
