If the roots of the equation 2x^2-x+6=0 are alpha and beta, find the eqn whose roots are?

For quadratic equation ax² + bx + c = 0

sum of roots = −b/a

product of roots = c/a

2x² − x + 6 = 0 has roots α and β

α + β = −(−1)/2 = 1/2

αβ = 6/2 = 3

(i)

Roots: 2α and 2β

−b/a = 2α + 2β = 2(α + β) = 2(1/2) = 1 ----> b = −a

c/a = 2α * 2β = 4αβ = 4(3) = 12 ----> c = 12a

For a, choose smallest positive integer that will make b and c integers:

a = 1, b = −1, c = 12

x² − x + 12 = 0

(ii)

Roots are 1/α and 1/β

−b/a = 1/α + 1/β = (α+β)/(αβ) = (1/2)/3 = 1/6 ---> b = −a/6

c/a = 1/α * 1/β = 1/(αβ) = 1/3 ---> c = a/3

a = 6, b = −1, c = 2

6x² − x + 2 = 0

(iii)

Roots are α+2β and β+2α

−b/a = (α+2β) + (β+2α) = 3(α + β) = 3/2 ----> b = −3a/2

c/a = (α+2β) (β+2α) = 2(α+β)² + αβ = 2(1/2)² + 3 = 7/2 ---> c = 7a/2

a = 2, b = −3, c = 7

2x² − 3x + 7 = 0

Define f(x) = 2x^2 -x +6.

(i) Horizontal dilation by a factor of 2 will double the value of each of the roots.

.. f(x/2) = 0 = 2(x/2)^2 -(x/2) +6 = x^2/2 -x/2 +6

.. 0 = x^2 -x +12 . . . . . multiply by 2 to eliminate fractions

(ii) f(1/x) = 0 will have roots that are the reciprocals of those of f(x).

.. f(1/x) = 0 = 2(1/x)^2 -(1/x) +6

.. 0 = 2 -x +6x^2 . . . . . multiply by x^2 to eliminate fractions

.. 0 = 6x^2 -x +2 . . . . . rearrange to make powers descending

(iii) Reflection of the function across the line x=α +β will transform the roots in the desired way. The value of α +β is -b/a, so is -(-1)/2 = 1/2.

.. f(2(α +β) -x) = f(1 -x)

.. f(1 -x) = 0 = 2(1 -x)^2 -(1 -x) +6 = 2 -4x +2x^2 -1 +x +6

.. 0 = 2x^2 -3x +7