Find k For Quadratic Equation?

Find k such that the equation x^2 - kx + 4 = 0 has a repeated real solution.

8 Answers

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  • 6 months ago
    Best Answer

    x^2 - kx + 4 = 0

    apply the discriminant formula

    a = 1 , b = - k, c = 4

    b^2 - 4ac = 0

    (-k)^2 - 4(1)(4) = 0

    k^2 - 16 = 0

    k^2 = 16

    k = √(16)

    k = 4 answer//

    x^2 - 4x + 4 = 0

    (x - 2)(x - 2) = 0

    x = 2

  • kv2=4ac

    k2= 4*1*4

    = 16

    k= 4

  • Como
    Lv 7
    6 months ago

    :-

    For equal roots : -

    k² = 4 a c

    k² = 4 x 1 x 4 = 16

    k = ± 4

  • 6 months ago

    We want the determinant to be 0.

    b^2 - 4ac = 0

    (-k)^2 - 4(1)(4) = 0

    k^2 = 16

    k = 4 or k = -4

    If k = 4, x = 2 with multiplicity 2.

    If k = -4, x = -2 with multiplicity 2.

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  • 6 months ago

    To form a square term which gives the repeat real solution you use

    k = 2√([x²][x⁰]) = 2√(1×4) = 2×2 = 4

    Note that [x²] is strictly positive and [x⁰] is strictly non-negative.

    x² - 4x + 4 = 0 is (x - 2)² = 0 with the repeated real solution x = 2.

  • alex
    Lv 7
    6 months ago

    ax^2+bx+c = 0 has a repeated real solution if b^2 - 4ac = 0

  • 6 months ago

    (x - a)(x - a) = x^2 - kx + 4

    x^2 - 2ax + a^2 = x^2 - kx + 4

    -2ax + a^2 = -kx + 4

    a = 2

    k = 4

  • rotchm
    Lv 7
    6 months ago

    A repeated real solution implies that it can be written as a(x-r)² . Expanding gives ax² -2arx + r². This must equal your given equation.

    From the cooef of x² we see that a=1. Can you see what r can be? So what is k?

    Or another way: a repeated real solution means that the discriminant "b² - 4ac" = 0. What can you conclude from that?

    Show your work/answers here & we can proceed if need be.

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