To solve for [tex]\( k \)[/tex] such that the polynomial [tex]\( 3x^2 - 11x + k \)[/tex] is a perfect square, we will follow the logic that a perfect square trinomial can be expressed in the form [tex]\((ax + b)^2\)[/tex].
1. Assume the Polynomial is a Perfect Square:
If [tex]\( 3x^2 - 11x + k \)[/tex] is a perfect square, it should be expressible as [tex]\((ax + b)^2\)[/tex].
2. Expand the Perfect Square:
Expanding [tex]\((ax + b)^2\)[/tex], we get:
[tex]\[
(ax + b)^2 = a^2x^2 + 2abx + b^2
\][/tex]
3. Match Coefficients with the Given Polynomial:
Compare this expanded form with the polynomial [tex]\( 3x^2 - 11x + k \)[/tex]:
[tex]\[
3x^2 - 11x + k = a^2x^2 + 2abx + b^2
\][/tex]
4. Determine [tex]\( a \)[/tex]:
From the coefficient of [tex]\( x^2 \)[/tex], we have:
[tex]\[
a^2 = 3
\][/tex]
Solving for [tex]\( a \)[/tex], we get:
[tex]\[
a = \sqrt{3} \quad \text{or} \quad a = -\sqrt{3}
\][/tex]
5. Determine [tex]\( b \)[/tex]:
From the coefficient of [tex]\( x \)[/tex], we have:
[tex]\[
2ab = -11
\][/tex]
Substituting [tex]\( a = \sqrt{3} \)[/tex], we solve for [tex]\( b \)[/tex]:
[tex]\[
2\sqrt{3}b = -11 \quad \Rightarrow \quad b = \frac{-11}{2\sqrt{3}} = -\frac{11\sqrt{3}}{6}
\][/tex]
6. Determine [tex]\( k \)[/tex]:
From the constant term, [tex]\( b^2 \)[/tex], we have:
[tex]\[
k = b^2
\][/tex]
Substituting the value of [tex]\( b \)[/tex]:
[tex]\[
b = -\frac{11\sqrt{3}}{6}
\][/tex]
Therefore:
[tex]\[
k = \left( -\frac{11\sqrt{3}}{6} \right)^2 = \frac{121 \cdot 3}{36} = \frac{363}{36} = \frac{121}{12}
\][/tex]
To summarize, the value of [tex]\( k \)[/tex] that makes the polynomial [tex]\( 3x^2 - 11x + k \)[/tex] a perfect square is:
[tex]\[
k = \frac{121}{12}
\][/tex]
