Answer :
Let's address the first part of the question regarding the roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] of the equation [tex]\(3x^2 - x - 1 = 0\)[/tex].
1. Identifying the roots, [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
The roots of the quadratic equation [tex]\(3x^2 - x - 1 = 0\)[/tex] are given as:
[tex]\[ \alpha = \frac{1}{6} - \frac{\sqrt{13}}{6} \][/tex]
[tex]\[ \beta = \frac{1}{6} + \frac{\sqrt{13}}{6} \][/tex]
2. Computing [tex]\(\frac{\alpha}{\beta}\)[/tex]:
[tex]\[ \frac{\alpha}{\beta} = \frac{\frac{1}{6} - \frac{\sqrt{13}}{6}}{\frac{1}{6} + \frac{\sqrt{13}}{6}} \][/tex]
3. Computing [tex]\(\frac{\beta}{\alpha}\)[/tex]:
[tex]\[ \frac{\beta}{\alpha} = \frac{\frac{1}{6} + \frac{\sqrt{13}}{6}}{\frac{1}{6} - \frac{\sqrt{13}}{6}} \][/tex]
So, the values of [tex]\(\alpha\)[/tex], [tex]\(\beta\)[/tex], [tex]\(\frac{\alpha}{\beta}\)[/tex], and [tex]\(\frac{\beta}{\alpha}\)[/tex] are:
[tex]\[ \alpha = \frac{1}{6} - \frac{\sqrt{13}}{6}, \quad \beta = \frac{1}{6} + \frac{\sqrt{13}}{6}, \quad \frac{\alpha}{\beta} = \frac{\frac{1}{6} - \frac{\sqrt{13}}{6}}{\frac{1}{6} + \frac{\sqrt{13}}{6}}, \quad \frac{\beta}{\alpha} = \frac{\frac{1}{6} + \frac{\sqrt{13}}{6}}{\frac{1}{6} - \frac{\sqrt{13}}{6}} \][/tex]
Now let's move to the second part of the question regarding the roots of the equation [tex]\(2x^2 + 3x - 4 = 0\)[/tex]:
1. Identifying the roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
For the quadratic equation [tex]\(2x^2 + 3x - 4 = 0\)[/tex], let's find the equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex].
Using the quadratic formula, the roots of [tex]\(ax^2 + bx + c = 0\)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\(2x^2 + 3x - 4 = 0\)[/tex]:
[tex]\[ \alpha, \beta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4} \][/tex]
2. Sum and product of the original roots:
The sum of the roots [tex]\(\alpha + \beta\)[/tex] is given by:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{3}{2} \][/tex]
The product of the roots [tex]\(\alpha \beta\)[/tex] is given by:
[tex]\[ \alpha \beta = \frac{c}{a} = -\frac{4}{2} = -2 \][/tex]
3. Sum and product of the new roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
The new roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex].
The sum of the new roots [tex]\(\alpha^2 + \beta^2\)[/tex] can be found using:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = \left(-\frac{3}{2}\right)^2 - 2(-2) = \frac{9}{4} + 4 = \frac{9}{4} + \frac{16}{4} = \frac{25}{4} \][/tex]
The product of the new roots [tex]\(\alpha^2 \beta^2\)[/tex] is:
[tex]\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = (-2)^2 = 4 \][/tex]
4. Forming the new quadratic equation:
The new quadratic equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex] is given by:
[tex]\[ x^2 - (\alpha^2 + \beta^2)x + \alpha^2 \beta^2 = 0 \][/tex]
Substituting the values we found:
[tex]\[ x^2 - \left(\frac{25}{4}\right)x + 4 = 0 \][/tex]
To clear the fraction, multiply through by 4:
[tex]\[ 4x^2 - 25x + 16 = 0 \][/tex]
So, the equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex] is:
[tex]\[ 4x^2 - 25x + 16 = 0 \][/tex]
1. Identifying the roots, [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
The roots of the quadratic equation [tex]\(3x^2 - x - 1 = 0\)[/tex] are given as:
[tex]\[ \alpha = \frac{1}{6} - \frac{\sqrt{13}}{6} \][/tex]
[tex]\[ \beta = \frac{1}{6} + \frac{\sqrt{13}}{6} \][/tex]
2. Computing [tex]\(\frac{\alpha}{\beta}\)[/tex]:
[tex]\[ \frac{\alpha}{\beta} = \frac{\frac{1}{6} - \frac{\sqrt{13}}{6}}{\frac{1}{6} + \frac{\sqrt{13}}{6}} \][/tex]
3. Computing [tex]\(\frac{\beta}{\alpha}\)[/tex]:
[tex]\[ \frac{\beta}{\alpha} = \frac{\frac{1}{6} + \frac{\sqrt{13}}{6}}{\frac{1}{6} - \frac{\sqrt{13}}{6}} \][/tex]
So, the values of [tex]\(\alpha\)[/tex], [tex]\(\beta\)[/tex], [tex]\(\frac{\alpha}{\beta}\)[/tex], and [tex]\(\frac{\beta}{\alpha}\)[/tex] are:
[tex]\[ \alpha = \frac{1}{6} - \frac{\sqrt{13}}{6}, \quad \beta = \frac{1}{6} + \frac{\sqrt{13}}{6}, \quad \frac{\alpha}{\beta} = \frac{\frac{1}{6} - \frac{\sqrt{13}}{6}}{\frac{1}{6} + \frac{\sqrt{13}}{6}}, \quad \frac{\beta}{\alpha} = \frac{\frac{1}{6} + \frac{\sqrt{13}}{6}}{\frac{1}{6} - \frac{\sqrt{13}}{6}} \][/tex]
Now let's move to the second part of the question regarding the roots of the equation [tex]\(2x^2 + 3x - 4 = 0\)[/tex]:
1. Identifying the roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
For the quadratic equation [tex]\(2x^2 + 3x - 4 = 0\)[/tex], let's find the equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex].
Using the quadratic formula, the roots of [tex]\(ax^2 + bx + c = 0\)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\(2x^2 + 3x - 4 = 0\)[/tex]:
[tex]\[ \alpha, \beta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4} \][/tex]
2. Sum and product of the original roots:
The sum of the roots [tex]\(\alpha + \beta\)[/tex] is given by:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{3}{2} \][/tex]
The product of the roots [tex]\(\alpha \beta\)[/tex] is given by:
[tex]\[ \alpha \beta = \frac{c}{a} = -\frac{4}{2} = -2 \][/tex]
3. Sum and product of the new roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
The new roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex].
The sum of the new roots [tex]\(\alpha^2 + \beta^2\)[/tex] can be found using:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = \left(-\frac{3}{2}\right)^2 - 2(-2) = \frac{9}{4} + 4 = \frac{9}{4} + \frac{16}{4} = \frac{25}{4} \][/tex]
The product of the new roots [tex]\(\alpha^2 \beta^2\)[/tex] is:
[tex]\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = (-2)^2 = 4 \][/tex]
4. Forming the new quadratic equation:
The new quadratic equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex] is given by:
[tex]\[ x^2 - (\alpha^2 + \beta^2)x + \alpha^2 \beta^2 = 0 \][/tex]
Substituting the values we found:
[tex]\[ x^2 - \left(\frac{25}{4}\right)x + 4 = 0 \][/tex]
To clear the fraction, multiply through by 4:
[tex]\[ 4x^2 - 25x + 16 = 0 \][/tex]
So, the equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex] is:
[tex]\[ 4x^2 - 25x + 16 = 0 \][/tex]
