To determine the value of [tex]\( k \)[/tex] for which the quadratic equation [tex]\((k-12) x^2 + 2(k-12)x + 2 = 0\)[/tex] has two real equal roots, we need to 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]. The discriminant ([tex]\( \Delta \)[/tex]) of this equation is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
For the given quadratic equation [tex]\((k-12)x^2 + 2(k-12)x + 2 = 0\)[/tex]:
1. Identify the coefficients:
- [tex]\( a = (k-12) \)[/tex]
- [tex]\( b = 2(k-12) \)[/tex]
- [tex]\( c = 2 \)[/tex]
2. Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex] into the discriminant formula:
[tex]\[
\Delta = [2(k-12)]^2 - 4[(k-12)] \cdot 2
\][/tex]
3. Simplify the terms inside the discriminant:
[tex]\[
\Delta = [2(k-12)]^2 - 8(k-12)
\][/tex]
[tex]\[
\Delta = 4(k-12)^2 - 8(k-12)
\][/tex]
4. For the roots to be real and equal, the discriminant must be zero:
[tex]\[
4(k-12)^2 - 8(k-12) = 0
\][/tex]
5. Factor the equation:
[tex]\[
4(k-12) [(k-12) - 2] = 0
\][/tex]
6. Set each factor to zero and solve for [tex]\( k \)[/tex]:
[tex]\[
4(k-12) = 0 \quad \text{or} \quad (k-12) - 2 = 0
\][/tex]
For the first factor:
[tex]\[
k-12 = 0
\][/tex]
[tex]\[
k = 12
\][/tex]
For the second factor:
[tex]\[
k-12 - 2 = 0
\][/tex]
[tex]\[
k - 14 = 0
\][/tex]
[tex]\[
k = 14
\][/tex]
7. Therefore, there are two values of [tex]\( k \)[/tex] for which the quadratic equation has two real and equal roots:
[tex]\[
k = 12 \quad \text{and} \quad k = 14
\][/tex]
Thus, [tex]\( k = 12 \)[/tex] and [tex]\( k = 14 \)[/tex] are the values for which the quadratic equation [tex]\((k-12)x^2 + 2(k-12)x + 2 = 0\)[/tex] has two equal real roots.
