maths question (1 Viewer)

[jellybelly59]

LOL... *sigh* can't do this one! It's a sign that i'm locked on holiday mode LOL! hope u guys could help me out with this one. probably too easy for u all

The question is : For what value of n will one root of the equation
(n-2)x^2 + (n+2)x + 2n + 1 = 0 be the reciprocal of the other?

 

[midifile]

LOL... *sigh* can't do this one! It's a sign that i'm locked on holiday mode LOL! hope u guys could help me out with this one. probably too easy for u all

The question is : For what value of n will one root of the equation
(n-2)x^2 + (n+2)x + 2n + 1 = 0 be the reciprocal of the other?

Let roots be a, 1/a

a x 1/a = (2n+1)/(n-2) (product of roots)
therefore (2n+1)/(n-2) = 1
n-2 = 2n +1
n = -3

 

[jellybelly59]

ty

 

[lyounamu]

Is it just me who thinks that n=-3 cannot be the solution?

if n=-3, the discriminant of the equation is less than 0 meaning that it doesn't even have roots to begin with.

 

[Aerath]

Yeah, if n = -3, discriminant is <0. But essentially midifile's way of doing it would've been the way I would've done it.

 

[lyounamu]

Yeah, if n = -3, discriminant is <0. But essentially midifile's way of doing it would've been the way I would've done it.

I did that and I realised that it didn't work so I think I need to use the addition one instead of the product one.

But I reckon the question might have been wrong. I certainly don't think question will ask a question that ask you to find roots when it doesn't even have roots in the first place.

 

[jaychouf4n]

If the question doesnt specify that the root is real then i think its ok to have a negative discriminant.

It just means it has a complex root.

 

[lyounamu]

If the question doesnt specify that the root is real then i think its ok to have a negative discriminant.

It just means it has a complex root.

Yeah, I know that. I am already aware that you can have unreal roots but this is the level of Year 11 Mathematics. There are cases where people may go back and check their answers. Hence they may wonder if they got their answers right or not.

I mean, that's the way I reassure that I got my questions right.

This is not even Mathematics Extension 2. Some people have no idea of what complex number is like, you know.

 

[midifile]

Is it just me who thinks that n=-3 cannot be the solution?

if n=-3, the discriminant of the equation is less than 0 meaning that it doesn't even have roots to begin with.

Yeah. I realised that when I had done it, but I dont think there is anything else that makes the question right.

I think that it the question is just flawed

 

[bored of sc]

(n-2)x² + (n+2)x + 2n + 1 = 0

Let roots be a and 1/a.

a + 1/a = -b/a = -(n+2)/(n-2) --------- (1)

a(1/a) = c/a = (2n+1)/(n-2) --------- (2)

From (2)

(2n+1)/(n-2) = a(1/a) = 1
2n + 1 = n - 2
n = -3

Sub n=-3 into (1)

a + 1/a = -(3+2)/(3-2)
a +1/a = -5
a² + 5a + 1 = 0

a = [-5+ square root(5² - 4x1x1)]/2
Sub a = [-5+ square root(21)]/2 back into (1)

-5 = -(n+2)/(n-2)
-5n+10 = -n-2
-4n = -12
n = 3

Therefore n = + 3

Jellybelly, is that the answer?

 

[lyounamu]

(n-2)x² + (n+2)x + 2n + 1 = 0

Let roots be a and 1/a.

a + 1/a = -b/a = -(n+2)/(n-2) --------- (1)

a(1/a) = c/a = (2n+1)/(n-2) --------- (2)

From (2)

(2n+1)/(n-2) = a(1/a) = 1
2n + 1 = n - 2
n = -3

Sub n=-3 into (1)

a + 1/a = -(3+2)/(3-2)
a +1/a = -5
a² + 5a + 1 = 0

a = [-5+ square root(5² - 4x1x1)]/2
Sub a = [-5+ square root(21)]/2 back into (1)

-5 = -(n+2)/(n-2)
-5n+10 = -n-2
-4n = -12
n = 3

Therefore n = + 3

Jellybelly, is that the answer?

That question is either flawed or he wrote a wrong question. There is no solution to that question.

 

[Just.Snaz]

Nah Namu, the question is right... prior to 4 unit maths, it is known that if the discriminant is negative, the roots are unreal (whatever that may mean) - but it still means there are roots

 

[jellybelly59]

solution at the back of couchman is n = -3

 

[Aerath]

Yeah, just use midifile's method.

 

[lsam]

Nah Namu, the question is right... prior to 4 unit maths, it is known that if the discriminant is negative, the roots are unreal (whatever that may mean) - but it still means there are roots

ok, so there are roots that are unreal.

*get back to maths textbook to have a look*

 

[Slidey]

So guys, what's the square root of -1?

 

[lyounamu]

So guys, what's the square root of -1?

ahhh...stop teasing people who have got no knowledge of EX2 maths.....

 

[Slidey]

Haha. TBH I think everybody from 2unit advanced onwards should know and learn complex numbers.

I reckon the only reason they're restricted to 4unit is because they have the word 'complex' in the name, which is a bit of a misnomer.

 

[Aplus]

Imaginery sounds more fun

 

[Yamiyo]

Some of the questions on polynomials purposefully involve unreal/irrational things like that to prevent you from doing it a different way...
e.g. compare the questions: find the sum of the zeroes of a) x^2 + 2x - 3 and b) x^2 + 2x + 3
(it's more difficult to just solve it and add them up)