Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

math problem?

For how many positive integers k less than 2020 does the equation below have real solutions for x?

(2x+1)(x+k)=(3x–2)(x−k)

4 Answers

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  • Ian H
    Lv 7
    1 month ago

    (3x – 2)(x − k) - (2x + 1)(x + k) = x^2 – (5k + 3)x  + k = 0 

    Real x if y = (5k + 3)^2 - 4k = 25k^2 + 26k + 9 ≥ 0 

    https://www.wolframalpha.com/input/?i=y+%3D+25k%5E... 

    That graph shows that y = 25k^2 + 26k + 9 ≥ 0 is positive for all k 

    So, all positive integers k less than 2020 have real solutions for x 

    Examples:  

    k = 100, x ~ 502.801 or x ~ 0.198886 

    k = 1000, x ~ 5002.8 or x ~ 0.199888 

    k = 2000, x ~ 10002.8 or x ~ 0.199944 

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  • Pope
    Lv 7
    1 month ago

    It is a quadratic equation. Expand it and group like terms.

    (2x + 1)(x + k) = (3x - 2)(x - k)

    2x² + (2k + 1)x + k = 3x² + (-3k - 2)x + 2k

    x² + (-5k - 3)x + k = 0

    The equation has real solutions if and only if the discriminant is greater than or equal to zero.

    discriminant ≥ 0

    (-5k - 3)² - 4(1)(k) ≥ 0

    25k² + 30k + 9 - 4k ≥ 0

    25k² + 26k + 9 ≥ 0

    A second discriminant, that of 25k² + 26k + 9, is -224.

    Therefore, 25k² + 26k + 9 has no real roots.

    The leading coefficient of 25k² + 26k + 9 is positive.

    Therefore, 25k² + 26k + 9 > 0 for all real k.

    Thus the given equation must have real solutions for all real k.

    There are 2019 positive integers less than 220. That is your number, 2019.

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  • david
    Lv 7
    1 month ago

    (2x+1)(x+k) = (3x–2)(x−k)

    2x^2 + x + 2kx + k = 3x^2 - 2x - 3kx + 2k

    x^2 - 5kx - 3x + k = 0

    b = -(5k + 3)

    a = 1  .. c = k

    discriminant

      b^2 - 4ac =  (5k + 3)^2 - 4(1)(k)

      =  25k^2 + 30k + 9 - 4k

      =  25k^2 + 26k + 9

     ... when k is positive, the discriminant is always positive which means ther will always be real solutions

      ...  all k < 2020  will give real solutions for x

     . so 2019 posivie valies of K create real solutions for x

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  • Philip
    Lv 6
    1 month ago

    k is an integer. 0 < k < 2020.

    (3x-2)(x-k)-(2x+1)(x+k) = 0, ie.,

    x^2[3-2]+x[-2-3k-2k-1]+k[2-1] =0, ie., x^2-8kx+k=0

    Then 2x=8k(+/-)D, where D^2 = (8k)^2-4k. For

    real roots we require D^2(=/>)0, ie., require

    4k(16k-1)(=/>)0, which holds for all k in [1,2019].

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