Question

For what value of k, does the equation $[k{x}^2+(2k+6)x+16=0]$ have equal roots?

• A. $1$ and $9$
• B. $-9$ and $-1$
• C. $-1$ and $9$
• D. $-1$ and $-9$

Answer 1

The correct option is A $1$ and $9$

For equal roots $D=0$

$\Rightarrow (2k+6{)}^2-4\times (k\left)\right(16)=0$

$\Rightarrow (2k+6{)}^2=4\times (k\left)\right(16)$

$\Rightarrow 4(k+3{)}^2=64k$

$\Rightarrow k^2+6k+9=16k$

$\Rightarrow k^2-10k+9=0$

$\therefore (k-1)(k-9)=0$

$\Rightarrow k=1$ or $k=9$