Question

Find the values of $k$ for the following quadratic equation, so that they have two real and equal roots:$(k-12)x^2+2(k-12)x+2=0$

• A. $k=-12$
• B. $k=14$
• C. $k=-14$
• D. $k=12$

Answer 1

Answer: (B), (D)

$k=12$
$k=14$

Given Quadratic equation is $(k-12)x^2+2(k-12)x+2=0$

Now, given that the roots are equal and zero.

For a quadratic equation $a{x}^2+bx+c=0$ roots are real and equal if $D=b^2-4ac=0$.

Here,$a=(k-12),b=2(k-12),c=2$

Therefore, $D=b^2-4ac=0$

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

$\Rightarrow 4(k^2-2\times k\times 12+{12}^2)-8(k-12)=0$

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

$\Rightarrow k^2-26k+168=0$

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

$\Rightarrow k(k-14)-12(k-14)=0$

$\Rightarrow (k-14)(k-12)=0$

$\Rightarrow k=14$ and $k=12$