find the value of K for which the given equation has equal roots (k-12) x^2+2(k-12)x+2=0​

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find the value of K for which the given equation has equal roots (k-12) x^2+2(k-12)x+2=0​

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  1. Answer:

    The values of k for which the given equation has equal roots are 14 and 12.

    Step-by-step explanation:

    Given polynomial is;

    (k-12) x^2+2(k-12)x+2=0;

    here, a = (k-12);

    b = -2(k-12);

    c = 2

    For equal roots, we know a discriminant and that is;

    b² – 4ac = 0;

    Thus , (-2(k-12))² – 4(k-12)(2) = 0;

    4(k-12)²- 4(k-12)(2) = 0;

    First value of k will be;

    4(k-12){(k-12) – 2} = 0;

    (k-12) – 2 = 0 / 4(k-12);

    k – 12 – 2 = 0;

    k – 14= 0;

    k = 14.

    Second value of k will be;

    4(k-12){(k-12) – 2} = 0;

    4(k-12){(k-12) – 2} = 0 / {(k-12) – 2};

    4(k-12) = 0;

    k – 12= 0;

    k = 12.

    That’s all.

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