grabthebasics.com
home faq wannahelp feedback contact

- solve quadratic on ti-83 plus
- forensic equations quadratic
- word problem solving using quadratic equation
- quadratic equations ti 89 titanium
- how to solve math using mdas
- solving word problems using quadratic equations
- solving demand equation
- quadratic equations can be solved by using the quadric equation formula
- introduction equations
- solving quadratic and cubic equations worksheets
- simplest method of solving quadratic equations
- quadriatic functions
- factorization of +quardratic equations
- wilkipedia+additional mathematics+quadratic equations
- step by step demand equations
- casio fx 3650p quatratic equation
- how to quadratic functions
- example of worded problems of quadratic equations
- partial differential equations simultaneous equations
- quadratic equation and real life

Song 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

home FAQ How can I help feedback Contact