Question

15th one.................​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Given Equation,

$\large{\sf{(k - 12)x {}^{2} - 2(k - 12)x \: + 2 = 0}}$

On comparison with ax² + bx + c = 0

$\sf{a = (k - 12) \: \: b = - 2(k - 12) \: and \: c = 2}$

Since,the zeros of the equation are equal

$\sf{\therefore, the \ discriminant \ is \ equal \ to \ zero}$

$\huge {\sf{d = b {}^{2} - 4ac = 0}}$

Putting the values,we get:

$\sf{[ - 2(k - 12)]{}^{2} - 4(k - 12)(2)} \: = 0 \\ \\ \rightarrow \: \sf{4(k - 12) {}^{2} - 8(k - 12) = 0} \\ \\ \rightarrow \: \sf{(k - 12) - 2 = 0} \\ \\ \huge{ \rightarrow \: \underline{ \boxed{ \sf{k = 14}}}}$

For k = 14,the given equation has real and equal roots

Answer 2

♠️$\huge\bold\green{HELLO}$♠️

Given Equation→

$\bold\blue{\tt{(k - 12)x {}^{2} - 2(k - 12)x \: + 2 = 0}}$

On comparison with ax² + bx + c = 0

$\tt{a = (k - 12) \: \: b = - 2(k - 12) \: and \: c = 2}$

Since,the zeros of the equation are equal

$\tt{\therefore, the \ discriminant \ is \ equal \ to \ zero}$

$\bold {\tt{d = b {}^{2} - 4ac = 0}}$

Putting the values,we get:

$\begin{lgathered}\tt{[ - 2(k - 12)]{}^{2} - 4(k - 12)(2)} \: = 0 \\ \\ \rightarrow \: \tt{4(k - 12) {}^{2} - 8(k - 12) = 0} \\ \\ \rightarrow \: \tt{(k - 12) - 2 = 0} \\ \\ \bold{ \rightarrow \: \underline{ \boxed{ \tt{k = 14}}}}\end{lgathered}$

For k = 14,the given equation has real and equal roots