Quadratic Equations???

[Dillinger4]

This shoud be easy for some of you math buffs out there but im kind of new with
this math stuff..... well anyway. this math software i bought explains Quadratic Equations as so........

Ax2 + Bx + C = 0
where A = 1, B = 5, C = 6

then they say multiply AC and find the factors of the product.... so 2 + 3 = 5 = B

ok so now they have X2 + 2x + 3x + 6 = 0
ok fine but next it get confusing...

X(X+2)+ 3(X + 2) = 0 why is this broken up as so.

then they goto (X + 2) (X + 3) = 0
then finaly X = -2 or X = -3

1.) Is it normal to have two answers for a Quadratic Equation?

2.) How does (X + 2) (X + 3) = 0 equate to X = -2 or X = -3 ?

3.) Is there a less confusing way to do this???

thanks all.............

[Guv]

Factoring always seems to be taught for purposes of developing some understanding, but it is not the way that quadratics are usually solved. There is a formula they will teach you. If A*X^2 + B*X + C = 0 Then one root is ( -B + SquareRoot(B^2 - 4*A*C) ) / 2*A The other is ( -B - SquareRoot(B^2 - 4*A*C) ) / 2*A Try it with X^2 + 5*X + 6Quadratics always have two solutions, but sometimes the roots are complex numbers instead of ordinary numbers. They will plague you with this concept later, but I could give you a rough idea if you want to think about it now.

If (X+2)*(X+3) = 0
Then either (X+2) = 0 or (X+3) = 0 will make product zero.
If (X+2) = 0, then X must be negative. In fact it is -2.

To be very formal X+2 = 0 - 2 = -2 Add equals to equals, getting equals. X = - 2

I think the formula is easier than factoring, and it always works. Factoring is a problem if the roots are not integers, and is just about impossible if the roots are never ending decimal fractions like (3 + 2*SquareRoot(5) ).

I hope the above helps. Post again if it does not.

[Dillinger4]

yes i noticed your formula in a previous post but i wasnt too sure how to apply it....... woud this be correct?

Ax2 + Bx + C = 0 where A = 1, B = 5, C = 6

formula is ( -B + SquareRoot(B^2 - 4*A*C) ) / 2*A

so...... (-5 + sqr(5^2 - 4*1*6)) / 2 * 1

(-5 + sqr(25 - 24)) / 2 * 1

(-5 + sqr(1)) / 2 * 1

-6 /2*1 = -3

x = 3

would that be correct?
thanks bro....

[Sam Finch]

Hi, I'll try to explain the basics of quadratic equations, then move on to the formula.


A quadratic equation is an equation in the form

aX^2 + bX + c = 0


these can also be expressed in the form

a(x - p)(x-q) = 0 (as long as a is not zero)

where p and q are the solutions to the equation.

The reason p and q are the solutions is that in the second form we set x = p we get.

a(p - p)(p - q) = 0

as p - p = 0 and anything multiplied by 0 is 0 a(p - p)(p-q) = 0. So p is a solution.

if we say x = q we get

a(q - p)(q - q) = 0

which is true because q - q is 0.

So to solve a quatratic equation all we have to do is find p and q where

a(x - p)(x - q) = aX^2 + bX + c


and p and q are our solutions.


lets look at a simple example where a = 1

x^2 + bx + c = 0 = (x - p)(x - q)

if we multiply out the brackets on the right hand side it'll look a bit like the left hand side.


(x - p)(x - q)

= x(x - q) - p(x - q)

= (x^2 - qx) - (px - pq)

= X^2 - (p + q)x + pq

if we compare this to the original formula

x^2 + bx + c


we see that we can find p and q by equating

-b = p + q
c = pq


so we have to find 2 numbers that add together to make - b and multiply together to make c

so if we try your example

x^2 + 5x + 6 = 0

we just have to find 2 numbers that add together to make - 5 and multiply together to make 6


what about -2 and -3

-2 + -3 = -5
-2 * -3 = 6

so -2 and -3 are the solutions if we put them into the equation


(-2)^2 + 5(-2) + 6
= 4 - 10 + 6
= 0

(-3)^2 + 5(-3) + 6
= 9 - 15 + 6
= 0


and those are the answers.


what about if a isn't 1.

well we have

ax^2 + bx + c = 0

so let's divide both sides by a

(a/a)x^2 + (b/a)x + c/a = 0/a

a/a = 1 and 0/a = 0

so this is just like the simple type of equation.

Try this one.

3x^2 + 18x - 21 = 0


first divide both sides by 3

(3/3)x^2 +(18/3)x - (21/3) = 0/3

so

x^2 + 6x - 7 = 0

so we need 2 numbers that add to make -6 and multiply to make -7

what about 1 and -7

1 + -7 = -6
1 * -7 = -7

so 1 and -7 are the 2 solutions.

check those in the original equation


3*(1^2) + 18*1 - 21 = 0

so 1 is a solution



3*(-7^2) + 18*(-7) - 21

= 3*49 - 18*7 - 21

= 147 - 126 - 21 = 0

so - 7 is a solution.


I've just noticed the sun's coming up so I'd better go to bed, don't worry about the formula for now, Guv or I will explain it later, go through that lot and see if it makes sense, using the formula is easier than factoring, because if you're factoring you still have to guess at the 2 numbers, but you should understand the factoring method, It's more important that you understand how the equations work and why there are 2 solutions. The formula's just something you have to learn.

[[Digital-X-Treme]]

Ok. In answer to the question, an easier way of working out the solution to that particluar quadratic is as follows. The tutorial you are learning from seems to be overcomplicating things for newbies.

You have the equation x^2 + 2x + 3x + 6 = 0. First collect like terms...
x^2 + 2x + 3x + 6 = 0 becomes...
x^2 + 5x + 6 = 0

This is the genaral form of a quadratic equation. There are two possible solutions for a quadratic.

Notice how the LHS (Left hand side) of the expression equals the 0 on the RHS (Right hand side). We need to find two factors (values involving x) that multiply together to make zero. Here is how we find them.

An easy and efficient way of finding solutions is to factorise a quadratic. This involves bracketing expressions that multiply out to give the original expression. To factorise our expression, we need to find two numbers that add to make 5 (the coefficient of x, or the B term) and multiply to make 6 (the C term). We know that to get x^2, we need to multiply x by x, so here goes.

Check that 2 and 3 work.

2 + 3 = 5 CORRECT
2 * 3 = 6 CORRECT

So, (x+2)*(x+3) = x^2 + 5x + 6 which we know equals 0

If the result of multiplying two factors is 0, then one of the values must be zero. This is the essence. There are TWO possible results.

Either (X+2) = 0 So, we solve this simple expression to find the first value of x.

x+2 = 0 (minus 2 from both sides)
x = -2

x+3 = 0 (minus 3 from both sides)
x = -3

So the two possible solutions for the equation x^2 + 5x + 6 are when x = -2, or x = -3.

Substituting these values into the original expression (replace x with the numerical values) will produce a result of 0.

There are numerous methods for solving quadratic equations. There is also a lot more stuff to know, as Guv and Sam have mentioned (such as imaginary solutions etc etc) If you require any help, post back.

I would also advise looking into an algebra text book so u can read about the theory of stuff, and not just 'see how to find the answer'. It is a lot easier trying to solve problems on an issue, if u know how it works. The software seems to be doing a bad job of explaining things.

Later

[[Digital-X-Treme]]

Sorry, i just saw your post in the middle asking if the solution was correct using the formula.

You have found one of the solutions, but there are two solutions. This occurs when you find the square root of a number, as two results are found, one posotive, and one negative. I.e. Sqr(4) = 2 or -2, as -2*-2 = 4 and 2*2 = 4.

This is how it goes...

x = -5 + Sqr(25-4*1*6) / 2*1
x = -5 + 1 / 2
x = -4 / 2
x = -2


x = -5 - Sqr(25-4*1*6) / 2*1
x = -5 - 1 / 2
x = -6 / 2
x = -3

Hope this of some help. Later.

[pranavdesai]

welll a quadratic equation can be generalised as

ax2 + bx + c
a quadratic eqn is of degree2 [degree being the maximum power that x has been raised to]
in case of a quadratic eqn the degree is 2.

while solving any eqn the no of ans u get is equal to the degree of the eqn
consider the ex
5x - 25 = 0
as u can see x has power 1 in 5*x
therefore this eqn will have just one solution
which can be given as follows

5x-25=0
adding 25 to both sides
5x-25+25= 0 + 25
5x = 25
therefore
x = 25/5
x = 5 only 1 soln
similrly a quadratic eqn being of degree 2 has two solns
consider
x2 + 5x + 6 = 0
here what u do is u compare the eqn with general eqn
ax2 + bx + c = 0
a=1
b=5
c=6

now the idea is to split up 6 in such a way that the two nos when multiplied result into 6
say 6 and 1
but the two nos should also add up to be equal to b which in this case is 5
if we take 6 and 1
then they add up to 7

then lets take 2 and 3
2*3=6
2+3=5
this is our required soln

therefore we split up
x2 + 5x + 6 = 0
as x2 + 3X + 2x + 6 = 0
now we take the common factors as per our requirement x(x+3) + 2(x+3) = 0
therefore
(x+3)(x+2) = 0
if m*n = 0
then either m=0
or n=o
similarly
either x + 3=0
or x+2 = 0
if x+3 = 0
x = -3 as explained in eqn with degree1

if x+2 = 0
x = -2

in addn to this std method and one more explainedabove
if u want an additional short formula u can pm me

[pranavdesai]

by far the most general method is
ax2 + bx + c = 0
then since its quadratic it will have two roots x1 and x2
formula for finding these roots is
x1= (-b - squareroot(b2 - 4ac))/2a
x2 = (-b + squareroot(b2-4ac))/2a

consider x2 - 4x + 4 = 0
here comparing this eqn with std eqn ax2 + bx + c =0
a =1
b = -4
c= 4

x1 and x2 can be found by substituting the values of a b and c in the above formula
x1 = (-(-4) -squareroot((-4)(-4) - 4*(1)(4)) / 2 * 1
here x1 =2
similarly using the formulae u can find x2= 2
if u havedoubts pm me