Question

Find $k$, so that $(k-12)x^2+2(k-12)x+2=0$ has equal roots, where $k\ne 12$

• A. $k=4$
• B. $k=12$
• C. $k=14$
• D. none of these

Answer 1

Answer: (C)

$k=14$

$ForequalrootsD=0$

${\left[2\right(k-12\left)\right]}^2-4.(k-12).2=0$

$4(k-12{)}^2-4(k-12)\times 2=0$

$k^2-24k+144-2k+24=0$

$or{k}^2-26k+168=0$

$k^2-14k-12k+168=0$

$k(k-14)-12(k-14)=0(k-12)(k-14)=0\therefore k=12or14\therefore k=14(ask\ne 12giveninquestion)$