Under what condition the sum of the roots of x^2 – mx + n = 0 is k times their
difference?​

By Mary

Under what condition the sum of the roots of x^2 – mx + n = 0 is k times their
difference?​

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Mary

1 thought on “Under what condition the sum of the roots of x^2 – mx + n = 0 is k times their<br />difference?​”

  1. Step-by-step explanation:

    Given:

    A quadratic equation is x^2 – mx + n = 0

    To find:

    Under what condition the sum of the roots of

    x^2 – mx + n = 0 is k times their difference?

    Solution:

    Given quadratic equation is x^2 – mx + n = 0

    On comparing with the standard quadratic equation ax^2 + bx + c = 0

    We have

    a = 1

    b= -m

    c = n

    Let α and β be the roots of the given equation

    (α > β)

    Sum of the roots = -b/a

    =>α + β = -(-m)/1

    =>α + β = m

    Sum of the roots = m————-(1)

    Product of the roots = c/a

    =>α × β = n/1

    =>αβ = n

    Diffrence of the roots = α – β

    We know that

    (a-b)^2=(a+b)^2-4ab

    =>(α – β)^2 = (α + β)^2 – 4αβ

    =>(α – β)^2 = m^2-4n

    =>α – β = √(m^2 -n)

    Difference between the roots = √(m^2 -n)—-(2)

    Sum of the roots = k× Difference of the roots

    From (1)&(2)

    =>m =K ×√(m^2 -n) (or)

    K = m/[√(m^2 -n)]

    Answer:

    The required condition for the given problem is

    m =K ×√(m^2 -n) (or) K = m/[√(m^2 -n)]

    Used formulae:

    • The standard quadratic equation is

    ax^2 + bx + c = 0

    • Sum of the roots = -b/a
    • Product of the roots = c/a
    • (a-b)^2=(a+b)^2-4ab
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