# Whats wrong with this mathematically?

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#### David

##### Deregistered
When solving a quadratic equation, such as this, we need to factorize first:

x^2+6x+8 = 0

=> (x+4)(x+2) = 0

So x = -4 or x = -2

Why can't we do x(x+6) = -8?
And so x = -8 or x = -6 (obviously wrong)

Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with... :embrass:

#### ~Arcanic~

##### Senior Member
Oh my... my maths is e kind of straight F9s... :sweat:

*this brings back nightmares....*

#### eikin

##### Senior Member
David said:
When solving a quadratic equation, such as this, we need to factorize first:

x^2+6x+8 = 0

=> (x+4)(x+2) = 0

So x = -4 or x = -2

Why can't we do x(x+6) = -8?
And so x = -8 or x = -6 (obviously wrong)

Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with... :embrass:
second equation is correct

in the second equation, x is still = -4 or -2

if x = -8

x(x+6) = -8(-8+6) = 16

if x = -6

x(x+6) = -6(-6+6) = 0

#### oqs

##### New Member
think the RHS has to be 0 because when 2 terms multiply with each other, (x+4)(x+2), the only way the result is 0 is that one of the term is zero. this way you can bring out each term and suppose that it is 0. that's why the 'or' in the answer.

however for x(x+6) = -8, x could be anything like -2, -1, -8... anything.. and (x+6) could be 4, 8, 1... anything, in accordance to the assumed value of x. You probably can find the answer too this way but i guess it's an exhausive search kind of thing. Hope this helps.

#### grantyale

##### Senior Member
hey,
from
x(x+a)=0
you can get x=0 or x=-a;
but there's no direct conclusion to be made if the right hand side is not ZERO...

#### roygoh

##### Senior Member
David said:
When solving a quadratic equation, such as this, we need to factorize first:

x^2+6x+8 = 0

=> (x+4)(x+2) = 0

So x = -4 or x = -2

Why can't we do x(x+6) = -8?
And so x = -8 or x = -6 (obviously wrong)

Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with... :embrass:
Multiply any number with zero and the outcome is zero. So if A * B = 0 then one or both of them must be zero.

As such, when a quadratic equation can be factored down to (x+4)(x+2) = 0, then either X+4 = 0 or X+2 = 0. So there are 2 solutions, X=-4 or X=-2.

If A * B = C and C<>0, then A=C only when B=1, or B=C only when A=1.

So if you re-express the euation in the form x(x+6) = -8, then it does not imply that X=-8 or X+6 = -8.

#### yaoxing

##### Senior Member
"ab = 0" iff "a = 0 or b = 0" for all real a, b.

i.e. if a product for two numbers is zero, then at least one of them must be zero. Likewise, like Roy mentioned, any product with a zero gives you zero.

Same reason cannot hold for a product that is non-zero because there are infinite ways to give a non-zero product involving real numbers.

There is a finite number of ways to give you a non-zero product IF you are restricted to only integer solutions though:

For example, ab = -8 iff
a = -8, b = 1 or
a = -4, b = 2 or
a = -2, b = 4 or
a = -1, b = 8 or
a = 1, b = -8 or
a = 2, b = -4 or
a = 4, b = -2 or
a = 8, b = -1.

If none of these combinations solve the given equation then we can say that the equation has no integer solution.

#### David

##### Deregistered
Wow thanks guys... quite a read!!!! Yup, makes sense.. :embrass:

#### +evenstar

##### Senior Member
David said:
Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with... :embrass:
This is because you're finding the roots of the equation, ie The x-intercepts. To find this, you'll have to equate y=o, and hence from the equation "y=ax^2+bx+c", you'll get "ax^2+bx+c=0" and you solve for x. Hope this helps =)

#### David

##### Deregistered
+evenstar said:
This is because you're finding the roots of the equation, ie The x-intercepts. To find this, you'll have to equate y=o, and hence from the equation "y=ax^2+bx+c", you'll get "ax^2+bx+c=0" and you solve for x. Hope this helps =)
Thanks evenstar... But I dun think that's the reason... I'll go with the ones given by the rest of the people above.

Anyway, thanks.

#### asturias105

##### Senior Member
David said:
Thanks evenstar... But I dun think that's the reason... I'll go with the ones given by the rest of the people above.

Anyway, thanks.
I think it was a sound reasoning.

#### +evenstar

##### Senior Member
asturias105 said:
I think it was a sound reasoning.
Err...it's how we do math in Seconday School. Others have higher educational level than me, hence they may have better knowledge than me...

#### Paul_Yeo

##### Senior Member
many many years didn't go maths liao :embrass:

#### slacker123

##### New Member
go draw out the graph of the equation you stated.

draw a line y = 8. you will find that this line will intersect at x=-6 and x=0.

as we want the "roots", the interesection of the curve with the x-axis. Naturally, we have let y=0.

#### sulhan

##### Moderator
Staff member
David said:
When solving a quadratic equation, such as this, we need to factorize first:

x^2+6x+8 = 0

=> (x+4)(x+2) = 0

So x = -4 or x = -2

Why can't we do x(x+6) = -8?
And so x = -8 or x = -6 (obviously wrong)

Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with... :embrass:
Solution where the curve cuts the x axis....
the discriminant...b^2-4ac tells you what type of solution you gonna get....
b^2-4ac > 0 two roots real and different -> this is yours x = -4 or x = -2
b^2-4ac = 0 two roots real and equal
b^2-4ac < 0 two roots which are maginary...

rgds,
sulhan

#### +evenstar

##### Senior Member
sulhan said:
Solution where the curve cuts the x axis....
the discriminant...b^2-4ac tells you what type of solution you gonna get....
b^2-4ac > 0 two roots real and different -> this is yours x = -4 or x = -2
b^2-4ac = 0 two roots real and equal
b^2-4ac < 0 two roots which are maginary...

rgds,
sulhan
Err...

b^2 - 4ac > 0 means 2 real roots (ie graph intercepts x-axis at 2 points)
b^2 - 4ac = 0 means 1 real root or 2 real roots which are equal (ie graph intercepts x axis at 1 point only)
b^2 - 4ac < 0 means no real roots (ie graph does not intercept x-axis)

#### yaoxing

##### Senior Member
I think David's question is, in essence, asking for the reason why
"ab = 0" lead to "a=0 or b=0"
but
"ab = c (where c is non-zero)" cannot give a conclusion the same way, i.e. lead to "a = c or b = c" or something similar. Not asking why we need to equate to zero instead of other things.

In fact, we CAN write an equation of the form x(x+6) = -8 and solve it like what slacker123 has described, i.e. draw the graphs of y = x(x+6) and y = -8 and see where they intersect. (But if we really wish to draw the graph, it makes more sense to draw y = x^2 + 6x + 8 and see where it intersects with y = 0, i.e. x-axis.)

Mathematically they yield the same solutions because the graph of y1 = x^2 +6x + 8 = (x+2)(x+4) is actually the graph of y2 = x(x+6) shifted upwards by 8-units. i.e. y1 = (x+2)(x+4) = x^2 + 6x + 8 = x(x+6) + 8 = y2 + 8. Thus, the x-coordinate where graph of y1 intersects y = 0 is the same as where y2 intersects y = -8, i.e 8 units below. Hope this should explain eikin's and slacker123's ideas.

#### user111

##### Senior Member
David said:
Why can't we do x(x+6) = -8?
And so x = -8 or x = -6 (obviously wrong)

step 1 : "x(x+6) = -8"

step 2: "hence x = -8 or x = -6"

because from step 1 to step 2, the logic is lost already.
i.e. step 2 is not a logical consequence of step 1. and how do we verify that from step 1 to step 2 is incorrect? simple. just subst x = -6 or x = -8 into x(x+6). u will never get -8

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