To find the value of [tex]\( k \)[/tex] for which the quadratic equation
[tex]\[
(k+1)x^2 - 6(k+1)x + 3(k+9) = 0, \quad k \neq -1
\][/tex]
has real and equal roots, we must ensure that the discriminant of the quadratic equation is zero.
The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex] and its discriminant [tex]\( \Delta \)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
For the roots to be real and equal, the discriminant must be equal to zero.
In our quadratic equation, the coefficients are:
[tex]\[
a = (k+1)
\][/tex]
[tex]\[
b = -6(k+1)
\][/tex]
[tex]\[
c = 3(k+9)
\][/tex]
The discriminant can be calculated as follows:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = [-6(k+1)]^2 - 4(k+1)[3(k+9)]
\][/tex]
First, let's compute [tex]\( [-6(k+1)]^2 \)[/tex]:
[tex]\[
[-6(k+1)]^2 = 36(k+1)^2
\][/tex]
Next, compute [tex]\( 4(k+1)[3(k+9)] \)[/tex]:
[tex]\[
4(k+1)[3(k+9)] = 12(k+1)(k+9)
\][/tex]
Now, substituting these into the discriminant expression:
[tex]\[
\Delta = 36(k+1)^2 - 12(k+1)(k+9)
\][/tex]
Setting the discriminant [tex]\( \Delta \)[/tex] to zero for real and equal roots:
[tex]\[
36(k+1)^2 - 12(k+1)(k+9) = 0
\][/tex]
Since [tex]\( k \neq -1 \)[/tex], factor out [tex]\( 12(k+1) \)[/tex]:
[tex]\[
12(k+1)[3(k+1)- (k+9)] = 0
\][/tex]
This simplifies to:
[tex]\[
12(k+1)(2k-6) = 0
\][/tex]
Since [tex]\( k \neq -1 \)[/tex], we focus on the factor [tex]\( 2k - 6 = 0 \)[/tex]:
[tex]\[
2k - 6 = 0
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
2k = 6
\][/tex]
[tex]\[
k = 3
\][/tex]
Therefore, the value of [tex]\( k \)[/tex] for which the quadratic equation has real and equal roots is [tex]\( \boxed{3} \)[/tex].
