Maths qns (algebra)

last SG · 2005-11-04T21:41:21+00:00

Sorry Star, if you were to ask me this question 7 years ago, I could have answered you perfectly... now I just feel blank Forgotten liauzz...

papercut87 SG

sinicker smart boy la..

sinicker SG

pls look at the shoutbox.

Justy SG

The only guy who bothered to attempt to help plus post his answer gets all this unwanted comments >.< Tsk tsk...

Btw, you're asking to factorise the equation, not solve right..?

The man who was death SG

yah loh wondering why so many ppl sovle it

kaobeikaobu SG

chim chim…very chim…

Gackt247 SG

Laughing

CoolMyth SG

Say 2x^2 - 3x - 5 = 0

2(x^2 - (3/2)x - 5/2) = 0

2(x^2 - (3/2)x + (3/4)^2 - 9/16 - 5/2) = 0

2(x^2 - (3/2)x + (3/4)^2) = 49/8

(x - (3/4))^2 = 49/16

Hence,

x - (3/4) = 7/4 or x - (3/4) = -7/4

x = 5/2 or x = - 1

The factorization you want is of the form:

(2x - 5)(x + 1) (since the initial polynomial has coefficient 2 for the highest degree)

vIn.Warrior SG

don't have to do "Completing the Square". Just have to apply the cross multiply method to obtain the answer. It'll be much faster and will not make as much mistake as when doing "Completing the Square"

CoolMyth SG

I'm just suggesting another method.

As my lecturer always say, you're doing maths here. Knowing one method is not enough. You have to know all possible methods before you can say that you are doing maths. Cool

launtpc SG

Nooo! Not solve, because behind the algebraic equation there is no "=0" so we cannot assume that.

We can only factorise.

x(2x-3)-5

Or can further factorise?

CoolMyth SG

By Euclidean Algorithm, for a map f: R -> R[x], x --> 2x^2 - 3x - 5, there exist g such that f = (x - a)g + r where a is a root, deg r < deg g.

It can be further proved that in this case that r = 0 since we know that f(x) = 2x^2 - 3x - 5 = (x + 1)(2x - 5) where a = 1, g = (2x - 5) and deg r = - infinity which is less than deg g = 1. Therefore the condition is satisfied and thus the factorization holds.

Though the "let f(x) = 0 and then solve" is not fundamentally correct but what's most important at this particular level is just to get 2x^2 - 3x - 5 = (x + 1)(2x - 5). It is definitely acceptable at this level.

Hope I didn't baffle anyone. Thanks for letting me have a small algebra revision. Laughing

renorenal SG

Fuih... so many maths expert here... potential actuarists Mr. Green

papercut87 SG

Shocked Shocked gong simi? lolz.. =p

CoolMyth SG

Maths lor. Laughing

sinicker SG

as i said earlier, mentally can liao.

nightzip SG

actually F(x)=0 is a boundary condition. Those studying Engineering should know this.

If we say F(x)=2x^2+3x-2, at most we can factorise only BUT cannot solve. Because we dont have the boundary condition. If FACT, F(x)=2x^2+3x-2 can have an INFINITE number of solutions, where F(x)=0 or F(x)=5, or F(x)=-1, etc... and the roots of the equation is normally DEFINED as when the equation of the curve CUTs the X-AXIS, i.e. y=f(x)=0, therefore, that is the boundary condition and therefore WHEN THE QUESTION STATES find the ROOTS of the equation, you can equate it equal to ZERO. Very Happy

hope that clears everything up.

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