grabthebasics.com
home faq wannahelp feedback contact

- algebrator simultaneous equations solving
- "online solver" + "simultaneous equation"
- how to +slove math solutions
- solutions to make quadratic equations
- "keane gonzales"
- lesson plans for solving quadratic equations
- quadratic equation problems difficult
- solving simultaneous equations step by step
- simultaneous equations basics
- quadratic equations in vertex form worksheet
- sample word problem in quadratic formula
- "simultanious equations"
- how to solve imaginary simultaneous equations with ti-86
- sample problems on quadratic equation
- lesson plans on graphing quadratics equations
- how to solve accounting equation
- solve for x 2 side equasions
- explaining quadratic functions data
- solving quadratic equations live
- ti-5 solving equation

What Is A True 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