F.4 MATHS QUAD EQUA.

1. Denote m = alpha, n = beta, m > n

And the roots differ by 2

So, m - n = 2

Sum of roots, m + n = -a

Product of roots, mn = b

m - n = 2

(m - n)^2 = 2^2 = 4

(m + n)^2 - 4mn = 4

(-a)^2 - 4b = 4

a^2 - 4b = 4

b = a^2/4 - 1



2. 2x^2 - 6x - 1 = 0

m + n = -(-6)/2 = 3

mn = -1/2

For the new equation, sum of roots = (m + 2n) + (2m + n) = 3(m + n) = 3(3) = 9

Product of roots = (m + 2n)(2m + n) = 2m^2 + 5mn + 2n^2

Since m and n are the roots of the equation,

2m^2 = 6m + 1 and 2n^2 = 6n + 1

So, Product of roots = (6m + 1) + 5mn + (6n + 1)

= 6(m + n) + 2 + 5mn

= 6(3) + 2 + 5(-1/2)

= 35/2

So, the new equation:

x^2 - 9x + 35/2 = 0

2x^2 - 18x + 35 = 0

1)α+β =-a
αβ =b
α-2=β
combined three equations
α+β =-a
α+α-2=-a
2α=2-a
α=(2-a)/2
then, αβ =b
β =b/α
β =(2b)/(2-a)

2)2x^2=6x+1
2x^2-6x-1=0
x^2-3x-1/2=0
α+β =3
αβ =-1/2
Thus, the required equation is,
[x-(α+2β)][x-(2α+β)]=0
x^2-3(α+β)x+ (α+2β)(2α+β)=0
x^2-3(α+β)x+( 2α^2+5αβ+2β^2) =0
x^2-3(α+β)x+ [2(α^2+β^2)+5αβ]=0
x^2-3(α+β)x+ {2[(α+β)^2-2αβ]+5αβ}=0
x^2-3(α+β)x+ 2[(α+β)^2-2αβ]+5αβ=0
x^2-3(3)x+ 2[(3)^2-2 (-1/2)]+5(-1/2)=0
x^2-9x+17.5=0
2x^2-18x+35=0