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### MC-II A Algebra Lesson 3, example question 3.5

MC-II A Algebra Lesson 3, example question 3.5

Hi

I did this question differently, instead of replacing the x with x-2, I just did the question as is, and when using the Vieta's theorem, taking into consideration that both roots are greater than 2, however, I got the wrong answer, I don't understand why, can someone help me?

My solution to this probem is as below:

First using discriminant >0, we get

X^2 + (m-2)x+5-m=0

disc= m^2 - 4m +4 - 4(5-m) = 0

m >=4 or m <=-4

Applying Vieta's theorem

X1 +X2=-b/a

and X1 and X2 are both >2, then

X1+X2 >4

so, 2-m >4 , which give us m<-2

X1*X2 = 5-m

and since X1 and X2 are both >2, then

x1*x2>4

so 5-m> 4, which give us m<1 ( this is the incorrect part, however, I don't understand why this is wrong?)

I understood how the teacher has done it in class, but don't under why my approach is incorrect, please help, thanks!

Re: MC-II A Algebra Lesson 3, example question 3.5

Hi Nicholas, good question! We need to be careful with some of the logic here.

Consider the two statements: (1) "X1 and X2 are both > 2" and (2) "X1+X2 > 4 and X1*X2 > 4".

Pretty clearly (1) implies (2), as if X1 and X2 are both > 2, then there sum and product are both > 4.

However, does (2) imply (1)? Here the answer is no. For example, if X1 = 1.5 and X2 = 4 then statement (2) is true but (1) is not.

This is what goes wrong with your method, as situations like the above X1 = 1.5 and X2 = 4 are not excluded.

What you really want to rely on here is positive and negative numbers, that is pos * pos = pos, pos * neg = neg, etc. This is where the idea of (X1-2) and (X2-2) comes from. If we look at a third statement (3) "(X1-2) + (X1-2) > 0 and (X1-2)(X2-2) > 0" we now do have that statements (1) and (3) are logically equivalent.

Hope this helps!