To solve the problem where one root of the quadratic equation [tex]\(5x^2 + 13x + k = 0\)[/tex] is the reciprocal of the other root, we can proceed as follows:
First, let the roots of the quadratic equation be [tex]\(r_1\)[/tex] and [tex]\(r_2\)[/tex]. According to the given condition, one root is the reciprocal of the other root, i.e., [tex]\(r_1 = \frac{1}{r_2}\)[/tex].
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], we know two key properties about its roots [tex]\(r_1\)[/tex] and [tex]\(r_2\)[/tex]:
1. The sum of the roots: [tex]\( r_1 + r_2 = -\frac{b}{a} \)[/tex]
2. The product of the roots: [tex]\( r_1 \cdot r_2 = \frac{c}{a} \)[/tex]
Given our specific quadratic equation [tex]\(5x^2 + 13x + k = 0\)[/tex]:
- [tex]\(a = 5\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = k\)[/tex]
Using the product of the roots:
[tex]\[ r_1 \cdot r_2 = \frac{c}{a} = \frac{k}{5} \][/tex]
Since [tex]\( r_1 = \frac{1}{r_2} \)[/tex], we substitute this into [tex]\( r_1 \cdot r_2 \)[/tex]:
[tex]\[ \frac{1}{r_2} \cdot r_2 = 1 \][/tex]
Therefore, we have:
[tex]\[ \frac{k}{5} = 1 \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = 5 \][/tex]
Thus, the value of [tex]\( k \)[/tex] is [tex]\(\boxed{5}\)[/tex].
