Question

For what value of k does (k-12)x^2+2(k-12)x+2=0 have equal roots

Answer 1

HELLO DEAR,

(k-12)x²+2(k-12)x+2=0

where, a=(k-12), b=2(k-12), c=2

have equal root

hence,

discriminant
$D = \sqrt{ {b}^{2} - 4ac} = 0$

now put the values of a,b,c

we get,

$\sqrt{ {(2k - 24})^{2} - 4 \times (k - 12) \times 2 } = 0 \\ = > \sqrt{4 {k}^{2} + {24}^{2} - 2 \times 24 \times 2k - 8k + 96 } = 0 \\ = > 4 {k}^{2} + 576 - 96k - 8k + 96 = 0.....squaring \: both \: side \\ = > 4 {k}^{2} - 104k + 672 = 0 \\ = > {k}^{2} - 26 + 168 = 0 \\ = > {k}^{2} - 12 - 14 + 168 = 0 \\ = > k(k - 12) - 14(k - 12) \\ = > (k - 12)(k - 14) = 0 \\ = > k = 12 \\ \\ \\ \\ \\ \: \: \: \: \: or \\ \\ \: \: \: \: \: k = 14$
I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

k equal to 12 or k equal to 14