Sum and product of Roots - Q ...:D (1 Viewer)

[Smile12345]

Hello All.

Could someone please help me with this Q?

Q. Find values of n in the equation 2x^2 - 5(n - 1)x + 12 = 0 if the two roots are consecutive.

Thanks for your help in advance.

 

[asianese]

Let the roots be @ and @+1.

 

[Smile12345]

Yup, I'd got that far...

So 2@ + 1 = n (sum) .... OR .... 2@ + @ = n (product)

 

[Rainbowtopp]

Are the answers n=(3+√137)/640 and n=(3-√137)/640?

 

[Smile12345]

Are the answers n=(3+√137)/640 and n=(3-√137)/640?

No... n = -1 and 3 (Apparently).

 

[Menomaths]

This is going to take ages to type up...

 

[Smile12345]

This is going to take ages to type up...

Oh sorry... Do you have time to do it? Thanks in advance.

 

[obliviousninja]

5(n-1)/2 = α + (α+1) Sum of roots
6 = α(α+1) Product of roots

Make α subject, sub into 1st equation.

 

[Smile12345]

5(n-1)/2 = α + (α+1) Sum of roots
6 = α(α+1) Product of roots

Solve simulataneously

Thanks heaps.. By this way I get n = -1... Now for the 2nd answer...

 

[Menomaths]

x²-5(n-1)x/2 + 6 = 0
alpha = a beta = a+1
a+a+1 = 2a+1
a(a+1) = a²+a
2a+1 = 5(n-1)/2
4a+2 = 5n-5
4a = 5n-7
a = (5n-7)/4
a²+a = 6
(5n-7/4)²+(5n-7)/4 = 6
(25n²-70n+49)/16 + (5n-7)/4 = 6
25n²-70n+49 + 20n-28 = 96
25²-50n-75 = 0
n= -1 or n= 3
Sorry it looks messy, but my keyboards messed up and I have to use an on-screen keyboard :'(

 

[Smile12345]

x²-5(n-1)x/2 + 6 = 0
alpha = a beta = a+1
a+a+1 = 2a+1
a(a+1) = a²+a
2a+1 = 5(n-1)/2
4a+2 = 5n-5
4a = 5n-7
a = (5n-7)/4
a²+a = 6
(5n-7/4)²+(5n-7)/4 = 6
(25n²-70n+49)/16 + (5n-7)/4 = 6
25n²-70n+49 + 20n-28 = 96
25²-50n-75 = 0
n= -1 or n= 3
Sorry it looks messy, but my keyboards messed up and I have to use an on-screen keyboard :'(

Thanks heaps Menomaths... No worries, thanks for taking the time.