Maths Question (1 Viewer)

[Dan895]

I had my half yearlies for MX1 today and there was a question that no one that I asked worked out. I dont think its particularly difficult however we are all stumped.

The question was a+b=6 and ab=4 find the values of a and b and solve 1/a + 1/b

 

[fortune_cookie]

a+b=6 -- eqn 1

ab=4 -- eqn 2

From eqn 1, a= 6-b

Sub into eqn 2

(6-b)b = 4

6b -b^2 = 4

b^2 -6b +4 =0

This solves to give b = 3+-sqrt(5)

When b=3+sqrt(5) , a= 6-b = 3-sqrt(5)

When b= 3-sqrt(5) , a = 6-b = 3+sqrt(5)

Therefore the two values of "a" and "b" are 3+sqrt(5) and 3-sqrt(5). It doesn't matter which one is which.


Therefore

1/a + 1/b = 1/[3+sqrt(5)] + 1/ [ 3-sqrt(5)]

= [ (3-sqrt(5) + (3+sqrt(5) ) ] / [ 3^2 -(sqrt(5))^2]

= 6/ [9-5]

= 3/2

 

[Fus Ro Dah]

I had my half yearlies for MX1 today and there was a question that no one that I asked worked out. I dont think its particularly difficult however we are all stumped.

The question was a+b=6 and ab=4 find the values of a and b and solve 1/a + 1/b

Very silly question imo, because you do not need the values of a and b to find the value of 1/a + 1/b.



The values are permutationally invariant due to commutativity of Addition and Multiplication in so it could go either way.

 

[VJ30]

I had my half yearlies for MX1 today and there was a question that no one that I asked worked out. I dont think its particularly difficult however we are all stumped.

The question was a+b=6 and ab=4 find the values of a and b and solve 1/a + 1/b

this was asked in maths extension 1???

 

[theind1996]

Very silly question imo, because you do not need the values of a and b to find the value of 1/a + 1/b.



The values are permutationally invariant due to commutativity of Addition and Multiplication in so it could go either way.

Haha yep. I don't see why no-one in OP's MX1 cohort worked that out..

 

[fortune_cookie]

Very silly question imo, because you do not need the values of a and b to find the value of 1/a + 1/b.



The values are permutationally invariant so it could go either way.

I didnt see that :|

I guess that is why the question said to specifically find the values. To force people to solve the equations. It's not a silly question when you think about it. If the student actually has to go through and solve the equations, it tests a fair bit in just one question.

Testing:

1. Simultaneous Equations.
2. Quadratic Formula.
3. Rationalising Denominators.

 

[Timske]

lol

 

[barbernator]

lol

lolwut

 

[theind1996]

I didnt see that :|

I guess that is why the question said to specifically find the values. To force people to solve the equations. It's not a silly question when you think about it. If the student actually has to go through and solve the equations, it tests a fair bit in just one question.

Testing:

1. Simultaneous Equations.
2. Quadratic Formula.
3. Rationalising Denominators.

Yea, but it's especially surprising (given that OP does MX1) that no-one in his cohort got it correct..

 

[barbernator]

Yea, but it's especially surprising (given that OP does MX1) that no-one in his cohort got it correct..

nobody he asked worked it out, he probably didnt ask everyone. I guess it is quite a worrying sign though