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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Eden 2 years 2021-07-17T15:58:50+00:00 1 Answers 0 views 0

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  1. Ava
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    2021-07-17T16:00:25+00:00
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    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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