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To solve for the value of [tex]\( k \)[/tex] given that one of the roots of the quadratic equation [tex]\( 3x^2 + kx + 1 = 0 \)[/tex] is the reciprocal of the other, follow these steps:
1. Let the roots of the equation be [tex]\( r \)[/tex] and [tex]\(\frac{1}{r}\)[/tex].
Since one root is [tex]\( r \)[/tex] and the other root is [tex]\(\frac{1}{r}\)[/tex], [tex]\( r \)[/tex] is not zero.
2. For a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the sum of the roots [tex]\( r_1 + r_2 \)[/tex] is given by [tex]\(-\frac{b}{a}\)[/tex] and the product of the roots [tex]\( r_1 \cdot r_2 \)[/tex] is given by [tex]\(\frac{c}{a}\)[/tex].
Comparing to our specific equation [tex]\( 3x^2 + kx + 1 \)[/tex]:
[tex]\[ a = 3, \quad b = k, \quad c = 1 \][/tex]
3. Sum of the roots:
[tex]\[ r + \frac{1}{r} = -\frac{k}{3} \][/tex]
4. Product of the roots:
[tex]\[ r \cdot \frac{1}{r} = \frac{1}{3} \][/tex]
[tex]\[ 1 = \frac{1}{3} \][/tex]
Now we are given one genuine product that must hold true, ensuring consistency:
[tex]\[ r \cdot \frac{1}{r} = 1 \][/tex]
5. Expand the quadratic equation using the roots:
Substituting [tex]\(r\)[/tex] and [tex]\(\frac{1}{r}\)[/tex] into the quadratic equation:
[tex]\[ 3(r^2) + kr + 1 = 0 \][/tex]
and for the reciprocal root:
[tex]\[ 3\left(\frac{1}{r^2}\right) + \frac{k}{r} + 1 = 0 \][/tex]
6. Simplify the quadratic equations:
First equation:
[tex]\[ 3r^2 + kr + 1 = 0 \][/tex]
Second equation:
[tex]\[ 3\left(\frac{1}{r^2}\right) + \frac{k}{r} + 1 = 0 \][/tex]
[tex]\[ \frac{3}{r^2} + \frac{k}{r} + 1 = 0 \][/tex]
7. Solve both equations by equating them to zero:
To obtain values for [tex]\(k\)[/tex]:
[tex]\[ \text{From the first root}:\, -3r - \frac{1}{r}= k \][/tex]
[tex]\[ \text{From the reciprocal root computation}:\, -r - \frac{3}{r} = k \][/tex]
Therefore, the possible values of [tex]\( k \)[/tex] that satisfy the conditions for the roots of the quadratic equation [tex]\( 3x^2 + kx + 1 = 0 \)[/tex] are:
[tex]\[ \boxed{[-3r - \frac{1}{r}, -r - \frac{3}{r}]} \][/tex]
1. Let the roots of the equation be [tex]\( r \)[/tex] and [tex]\(\frac{1}{r}\)[/tex].
Since one root is [tex]\( r \)[/tex] and the other root is [tex]\(\frac{1}{r}\)[/tex], [tex]\( r \)[/tex] is not zero.
2. For a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the sum of the roots [tex]\( r_1 + r_2 \)[/tex] is given by [tex]\(-\frac{b}{a}\)[/tex] and the product of the roots [tex]\( r_1 \cdot r_2 \)[/tex] is given by [tex]\(\frac{c}{a}\)[/tex].
Comparing to our specific equation [tex]\( 3x^2 + kx + 1 \)[/tex]:
[tex]\[ a = 3, \quad b = k, \quad c = 1 \][/tex]
3. Sum of the roots:
[tex]\[ r + \frac{1}{r} = -\frac{k}{3} \][/tex]
4. Product of the roots:
[tex]\[ r \cdot \frac{1}{r} = \frac{1}{3} \][/tex]
[tex]\[ 1 = \frac{1}{3} \][/tex]
Now we are given one genuine product that must hold true, ensuring consistency:
[tex]\[ r \cdot \frac{1}{r} = 1 \][/tex]
5. Expand the quadratic equation using the roots:
Substituting [tex]\(r\)[/tex] and [tex]\(\frac{1}{r}\)[/tex] into the quadratic equation:
[tex]\[ 3(r^2) + kr + 1 = 0 \][/tex]
and for the reciprocal root:
[tex]\[ 3\left(\frac{1}{r^2}\right) + \frac{k}{r} + 1 = 0 \][/tex]
6. Simplify the quadratic equations:
First equation:
[tex]\[ 3r^2 + kr + 1 = 0 \][/tex]
Second equation:
[tex]\[ 3\left(\frac{1}{r^2}\right) + \frac{k}{r} + 1 = 0 \][/tex]
[tex]\[ \frac{3}{r^2} + \frac{k}{r} + 1 = 0 \][/tex]
7. Solve both equations by equating them to zero:
To obtain values for [tex]\(k\)[/tex]:
[tex]\[ \text{From the first root}:\, -3r - \frac{1}{r}= k \][/tex]
[tex]\[ \text{From the reciprocal root computation}:\, -r - \frac{3}{r} = k \][/tex]
Therefore, the possible values of [tex]\( k \)[/tex] that satisfy the conditions for the roots of the quadratic equation [tex]\( 3x^2 + kx + 1 = 0 \)[/tex] are:
[tex]\[ \boxed{[-3r - \frac{1}{r}, -r - \frac{3}{r}]} \][/tex]
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