Question

If the eqn x²+2(k+2)x+9=0 has equal roots ,then find the value of k

Answer 1

given ,

x²+2(k+2)x+9=0

there , a = 1

b = 2(k+2)

c = 9

and the eqn x²+2(k+2)x+9=0 has equal roots

so ,

${ {b}^{2} } - 4ac = 0 \\ = >( {2(k + 2))}^{2} - 4 \times 1 \times 9 = 0 \\ = > 4 {(k + 2)}^{2} - 36 = 0 \\ = > 4( {k}^{2} + 4k + 4) - 36 = 0\\ = > 4 {k}^{2} + 16k + 16 - 36 = 0 \\ = > 4 {k}^{2} + 16k - 20 = 0 \\ = > 4 {k}^{2} + 20k - 4k - 20 = 0 \\ = > 4k(k + 5) - 4(k + 5) = 0 \\ = > (k + 5)((4k - 4) = 0 \\ so \: \: \\ \: \: \: \: \: \: \: \: k + 5 = 0 \: \: \: \: or \: \: \: \: \: 4k - 4 = 0 \\ \: \: \: \: \: \: \: \: k = - 5 \: \: \: \: \: \: \: \: \: = > k \: = \frac{4}{4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1$