Question

4 Find the value of K.
so that the root are equal
X2+ 7(3+2K) -2x(1+3k)=0.
equal..

Answer 1

Answer:

${x}^{2} - 2(1 + 3k)x + 7(3 + 2k) \\ disc = ( - 2(1 + 3k))^{2} - 4(1)(7(3 + 2k)) \\ = 4(1 + 9 {k}^{2} + 6k) - 28(3 + 2k) \\ = 4 + 36 {k}^{2} + 24k - 84 -5 6k \\ = 36 {k}^{2} - 32k - 80 \\ as \: roots \: are \: equal \: so \: disc = 0 \\ 4(9 {k}^{2} - 8k - 20) = 0 \\ 9 {k}^{2} - 18k + 10k - 20 = 0 \\ 9k(k - 2) + 10(k - 2) = 0 \\ (k - 2)(9k + 10) = 0 \\ k = 2 \: \: \: \: \: \: k = \frac{ - 10}{9}$