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#1 2007-02-15 11:56:06

roxxy_the_raven13
Member
Registered: 2007-02-15
Posts: 1

Quadratic Equations and Trinomials

Here is the problem x^2 + 5x -24

How in the world do I solve this?

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#2 2007-02-15 12:54:33

fcforte07
Member
Registered: 2007-02-03
Posts: 7

Re: Quadratic Equations and Trinomials

it's more of x²-3x-28=0, the answer in the back of the book says (7,-4) are the answers. We just want to know the process.

And then there's 4x²=3x with the answer (0,-3/4)
so far i got to 4x-3x=0 then i'm stuck. The example we got had the "factor" box thing to get the answer, but it's a different problem.

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#3 2007-02-15 14:50:13

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Quadratic Equations and Trinomials

Try Quadratic Equation Solver

And the explanation is on Quadratic Equation


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#4 2007-02-15 16:57:17

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 45,956

Re: Quadratic Equations and Trinomials

x²-3x-28=0

By factorization method,
=> x²-7x+4x-28=0

=> x(x-7)+4(x-7)=0

=> (x-7)(x+4)=0

=> (x-7)=0 or (x+4)=0 or both equal to zero.

Yherefore, x=7 or x=-4.

In factorization method, the term containing the variaible of degree 1, (x in this case) has to be split in such a way that their sum is equal to the coefficient of the 'x' term (-3 in the above case) and their product is equal to the coeffficient of the term without variable (-28 in the above case).


Similarly,
4x²=3x
=> 4x²-3x=0
=>x(4x-3)=0
=>x=0 or 4x=3
=>x=0 or x=3/4.

But the formula for solution of a Quadratic equation is always much easier to use.


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#5 2007-02-15 17:23:47

pi man
Member
Registered: 2006-07-06
Posts: 251

Re: Quadratic Equations and Trinomials

You need to break the equations into factors.   From experience, I know that the solution will probably involve breaking down the equation into the form (x+a) (x+b) = 0.   Let's see what happens when you multiply that equation out:

Your initial equation looks like that.   In this case, a+b = 5 and ab=-24.  Keep in mind that either or both a and b can be less than 0.   What 2 numbers multplied together equal 24?   There's the following pairs of numbers:  1 and -24, -1 and 24, 2 and -12, -2 and 12, 3 and -8,
-3 and 8, 4 and -6, -6 and 4.   

Now of those groups which, if any, of those pairs when added together equal 5?   Just one of them and it's  -3 and 8.   So your equation can be factored into (x-3)(x+8)=0 .   The only ways that equation can be equal to 0 is when x-3=0 or x+8=0  That happens when x is equal to 3 or -8.   So that's your answer.   Plug those values in for x to double-check your solution.   

Both solutions check out okay.   

Equations don't often factor out as easily as this one did.   Your book is giving you these questions because they do factor cleanly.   Later I'm sure you'll get equations that are as easy and that's when the quadratic formula comes in handy. 

Let's look at the second problem.   x²-3x-28=0.  The same method can be used.  You want to find a and b where ab=-28 and a+b=-3.  You'll find out that a and b work out to -7 and 4 so that you get (x-7)(x+4)=0.  Again, the only way that equation can be equal to 0 is if one of the 2 equations inside the parenthesis is equal to 0.   So x is equal to 7 or -4.  These types of problems will get easier with experience and practice.

The third problem is a little different.  4x²=3x 

Same prinicipal now.   For that equation to be equal to 0, either 4x-3=0 or x=0.     For 4x-3=0, x=3/4.

Edited - Ganesh beat me to the punch.   While the quadratic equation may sometimes be simpler and always work, I'm sure your teacher is looking for you to factor out these equations.   Personally I always took a quick look to see if it can be easily factored.  If so, I'm done.  If not, then I resort to the quadratic equation.

Last edited by pi man (2007-02-15 17:29:18)

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