Answer :
Let's solve the given problem step by step.
We start with the quadratic equation [tex]\(x^2 - px + q = 0\)[/tex], where [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots.
By Vieta's formulas, we know that:
1. [tex]\(\alpha + \beta = p\)[/tex]
2. [tex]\(\alpha \beta = q\)[/tex]
We need to form a new quadratic equation whose roots are [tex]\(\frac{\alpha}{\beta^3}\)[/tex] and [tex]\(\frac{\beta}{\alpha^3}\)[/tex].
First, let's determine the sum of these new roots. The new roots are:
[tex]\[ \frac{\alpha}{\beta^3} \quad \text{and} \quad \frac{\beta}{\alpha^3} \][/tex]
To find the sum of the new roots, we calculate:
[tex]\[ \frac{\alpha}{\beta^3} + \frac{\beta}{\alpha^3} \][/tex]
We can express this in terms of [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ \frac{\alpha}{\beta^3} + \frac{\beta}{\alpha^3} = \frac{\alpha^4 + \beta^4}{\alpha^3 \beta^3} \][/tex]
Since [tex]\(\alpha \beta = q\)[/tex], we have:
[tex]\[ \alpha^3 \beta^3 = (q)^3 = q^3 \][/tex]
Thus:
[tex]\[ \frac{\alpha^4 + \beta^4}{q^3} \][/tex]
Next, we need to find [tex]\(\alpha^4 + \beta^4\)[/tex]. Using the identity for the sum of fourth powers, we get:
[tex]\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \][/tex]
We utilize the fact that [tex]\((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha \beta\)[/tex] and rearrange it to find [tex]\(\alpha^2 + \beta^2\)[/tex]:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \][/tex]
[tex]\[ \alpha^2 + \beta^2 = p^2 - 2q \][/tex]
Now substitute this back into the sum of fourth powers:
[tex]\[ \alpha^4 + \beta^4 = (p^2 - 2q)^2 - 2q^2 \][/tex]
Expanding and simplifying, we get:
[tex]\[ \alpha^4 + \beta^4 = p^4 - 4p^2 q + 4q^2 - 2q^2 \][/tex]
[tex]\[ \alpha^4 + \beta^4 = p^4 - 4p^2 q + 2q^2 \][/tex]
Thus, the sum of the new roots is:
[tex]\[ \frac{p^4 - 4p^2 q + 2q^2}{q^3} \][/tex]
Next, we calculate the product of the new roots:
[tex]\[ \left(\frac{\alpha}{\beta^3}\right) \left(\frac{\beta}{\alpha^3}\right) = \frac{\alpha \beta}{\alpha^3 \beta^3} = \frac{1}{\alpha^2 \beta^2} \][/tex]
Considering [tex]\(\alpha \beta = q\)[/tex]:
[tex]\[ \alpha^2 \beta^2 = q^2 \][/tex]
Thus, the product of the new roots is:
[tex]\[ \frac{1}{q^2} \][/tex]
Finally, we form the new quadratic equation using these roots. If the roots of a quadratic equation are [tex]\(r\)[/tex] and [tex]\(s\)[/tex], then the equation is of the form:
[tex]\[ x^2 - (r+s)x + rs = 0 \][/tex]
Substituting the sum and the product of the new roots, the equation is:
[tex]\[ x^2 - \left(\frac{p^4 - 4p^2 q + 2q^2}{q^3}\right)x + \frac{1}{q^2} = 0 \][/tex]
Simplifying, we get:
[tex]\[ x^2 - \left(\frac{p^4 - 4p^2 q + 2q^2}{q^3}\right)x + q^{-2} = 0 \][/tex]
Thus, the quadratic equation whose roots are [tex]\(\frac{\alpha}{\beta^3}\)[/tex] and [tex]\(\frac{\beta}{\alpha^3}\)[/tex] is:
[tex]\[ x^2 - \left( \frac{p^4 - 4p^2 q + 2q^2}{q^3} \right)x + \frac{1}{q^2} = 0 \][/tex]
We start with the quadratic equation [tex]\(x^2 - px + q = 0\)[/tex], where [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots.
By Vieta's formulas, we know that:
1. [tex]\(\alpha + \beta = p\)[/tex]
2. [tex]\(\alpha \beta = q\)[/tex]
We need to form a new quadratic equation whose roots are [tex]\(\frac{\alpha}{\beta^3}\)[/tex] and [tex]\(\frac{\beta}{\alpha^3}\)[/tex].
First, let's determine the sum of these new roots. The new roots are:
[tex]\[ \frac{\alpha}{\beta^3} \quad \text{and} \quad \frac{\beta}{\alpha^3} \][/tex]
To find the sum of the new roots, we calculate:
[tex]\[ \frac{\alpha}{\beta^3} + \frac{\beta}{\alpha^3} \][/tex]
We can express this in terms of [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ \frac{\alpha}{\beta^3} + \frac{\beta}{\alpha^3} = \frac{\alpha^4 + \beta^4}{\alpha^3 \beta^3} \][/tex]
Since [tex]\(\alpha \beta = q\)[/tex], we have:
[tex]\[ \alpha^3 \beta^3 = (q)^3 = q^3 \][/tex]
Thus:
[tex]\[ \frac{\alpha^4 + \beta^4}{q^3} \][/tex]
Next, we need to find [tex]\(\alpha^4 + \beta^4\)[/tex]. Using the identity for the sum of fourth powers, we get:
[tex]\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \][/tex]
We utilize the fact that [tex]\((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha \beta\)[/tex] and rearrange it to find [tex]\(\alpha^2 + \beta^2\)[/tex]:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \][/tex]
[tex]\[ \alpha^2 + \beta^2 = p^2 - 2q \][/tex]
Now substitute this back into the sum of fourth powers:
[tex]\[ \alpha^4 + \beta^4 = (p^2 - 2q)^2 - 2q^2 \][/tex]
Expanding and simplifying, we get:
[tex]\[ \alpha^4 + \beta^4 = p^4 - 4p^2 q + 4q^2 - 2q^2 \][/tex]
[tex]\[ \alpha^4 + \beta^4 = p^4 - 4p^2 q + 2q^2 \][/tex]
Thus, the sum of the new roots is:
[tex]\[ \frac{p^4 - 4p^2 q + 2q^2}{q^3} \][/tex]
Next, we calculate the product of the new roots:
[tex]\[ \left(\frac{\alpha}{\beta^3}\right) \left(\frac{\beta}{\alpha^3}\right) = \frac{\alpha \beta}{\alpha^3 \beta^3} = \frac{1}{\alpha^2 \beta^2} \][/tex]
Considering [tex]\(\alpha \beta = q\)[/tex]:
[tex]\[ \alpha^2 \beta^2 = q^2 \][/tex]
Thus, the product of the new roots is:
[tex]\[ \frac{1}{q^2} \][/tex]
Finally, we form the new quadratic equation using these roots. If the roots of a quadratic equation are [tex]\(r\)[/tex] and [tex]\(s\)[/tex], then the equation is of the form:
[tex]\[ x^2 - (r+s)x + rs = 0 \][/tex]
Substituting the sum and the product of the new roots, the equation is:
[tex]\[ x^2 - \left(\frac{p^4 - 4p^2 q + 2q^2}{q^3}\right)x + \frac{1}{q^2} = 0 \][/tex]
Simplifying, we get:
[tex]\[ x^2 - \left(\frac{p^4 - 4p^2 q + 2q^2}{q^3}\right)x + q^{-2} = 0 \][/tex]
Thus, the quadratic equation whose roots are [tex]\(\frac{\alpha}{\beta^3}\)[/tex] and [tex]\(\frac{\beta}{\alpha^3}\)[/tex] is:
[tex]\[ x^2 - \left( \frac{p^4 - 4p^2 q + 2q^2}{q^3} \right)x + \frac{1}{q^2} = 0 \][/tex]
