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

(i) 2alpha, 2beta

(ii) 1/alpha, 1/beta

(iii) alpha +2beta, beta+2alpha

given answer are

(i) x^2-x+12=0

(ii) 6x^2-x+2=0

(iii) 2x^2-3x+7=0

2 Answers

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  • 3 years ago
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    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

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  • 3 years ago

    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

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