Question

Find the values of k for which the following equation has equal roots: [4 MARKS]

$(k-12)x^2+2(k-12)x+2=0$

Answer 1

Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

We have,

$(k-12)x^2+2(k-12)x+2=0$

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

$\therefore D=b^2-4ac=4(k-12{)}^2-4(k-12)\times 2$

$\Rightarrow D=4(k-12)\left\{\right(k-12)-2\}$

$\Rightarrow D=4(k-12)(k-14)$

The given equation will have equal roots, if

$D=0\Rightarrow 4(k-12)(k-14)=0$

$\Rightarrow k-12=0or,k-14=0$

$\Rightarrow k=12or,k=14$

But, if $k=12$, then the given equation will not be quadratic.
Hence, $k=14$ is the only solution.