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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!

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!