Discover a world of knowledge and community-driven answers at IDNLearn.com today. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.
Sagot :
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]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.
