If $$ \alpha $$ and $$ \beta $$ are the roots of $$ a x^{2}+b x+c=0 $$, find the equations whose roots are $$ \begin{array}{ll} ext { a. } \alpha \beta^{-1}, \beta \alpha^{-1} & ext { b. } \alpha^{2}, \beta^{2} \ ext { d. } \alpha^{3}, \beta^{3} & ext { e. } \sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}} \ ext { If the roots of the equation } a x^{2}+c x+c=0 & ext { are in the ratio } m: n ext {, then prove that }(a \alpha+b)^{-1},(a \beta+b\end{array} $$

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If $$ \alpha $$ and $$ \beta $$ are the roots of $$ a x^{2}+b x+c=0 $$, find the equations whose roots are $$ \begin{array}{ll}	ext { a. } \alpha \beta^{-1}, \beta \alpha^{-1} & 	ext { b. } \alpha^{2}, \beta^{2} \ 	ext { d. } \alpha^{3}, \beta^{3} & 	ext { e. } \sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}} \ 	ext { If the roots of the equation } a x^{2}+c x+c=0 & 	ext { are in the ratio } m: n 	ext {, then prove that }(a \alpha+b)^{-1},(a \beta+b\end{array} $$

UpStudy AI · Accepted Answer

To find the equations whose roots are transformations of the roots $$ \alpha $$ and $$ \beta $$ of the quadratic equation $$ ax^2 + bx + c = 0 $$, we will use Vieta's formulas, which state that for a quadratic equation $$ ax^2 + bx + c = 0 $$ with roots $$ \alpha $$ and $$ \beta $$:

- The sum of the roots $$ \alpha + \beta = -\frac{b}{a} $$
- The product of the roots $$ \alpha \beta = \frac{c}{a} $$

Now, let's find the required equations for each case:

### a. Roots $$ \alpha \beta^{-1}, \beta \alpha^{-1} $$

Let $$ r_1 = \alpha \beta^{-1} $$ and $$ r_2 = \beta \alpha^{-1} $$.

1. **Sum of the roots**:
   $$
   r_1 + r_2 = \alpha \beta^{-1} + \beta \alpha^{-1} = \frac{\alpha^2 + \beta^2}{\alpha \beta}
   $$
   Using the identity $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$:
   $$
   \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2 - 2ac}{a^2}
   $$
   Thus,
   $$
   r_1 + r_2 = \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c}{a}} = \frac{b^2 - 2ac}{ac}
   $$

2. **Product of the roots**:
   $$
   r_1 r_2 = \alpha \beta^{-1} \cdot \beta \alpha^{-1} = 1
   $$

The equation with roots $$ r_1 $$ and $$ r_2 $$ is:
$$
x^2 - \left(\frac{b^2 - 2ac}{ac}\right)x + 1 = 0
$$

### b. Roots $$ \alpha^2, \beta^2 $$

Let $$ r_1 = \alpha^2 $$ and $$ r_2 = \beta^2 $$.

1. **Sum of the roots**:
   $$
   r_1 + r_2 = \alpha^2 + \beta^2 = \frac{b^2 - 2ac}{a^2}
   $$

2. **Product of the roots**:
   $$
   r_1 r_2 = \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{c}{a}\right)^2
   $$

The equation with roots $$ r_1 $$ and $$ r_2 $$ is:
$$
x^2 - \left(\frac{b^2 - 2ac}{a^2}\right)x + \left(\frac{c}{a}\right)^2 = 0
$$

### d. Roots $$ \alpha^3, \beta^3 $$

Let $$ r_1 = \alpha^3 $$ and $$ r_2 = \beta^3 $$.

1. **Sum of the roots**:
   $$
   r_1 + r_2 = \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = \left(-\frac{b}{a}\right)\left(\frac{b^2 - 2ac}{a^2}\right) = -\frac{b(b^2 - 2ac)}{a^3}
   $$

2. **Product of the roots**:
   $$
   r_1 r_2 = \alpha^3 \beta^3 = (\alpha \beta)^3 = \left(\frac{c}{a}\right)^3
   $$

The equation with roots $$ r_1 $$ and $$ r_2 $$ is:
$$
x^2 + \frac{b(b^2 - 2ac)}{a^3}x + \left(\frac{c}{a}\right)^3 = 0
$$

### e. Roots $$ \sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}} $$

Let $$ r_1 = \sqrt{\frac{\alpha}{\beta}} $$ and $$ r_2 = \sqrt{\frac{\beta}{\alpha}} $$.

1. **Sum of the roots**:
   $$
   r_1 + r_2 = \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} = \frac{\alpha + \beta}{\sqrt{\alpha \beta}} = \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}} = -\frac{b}{\sqrt{ac}}
   $$

2. **Product of the roots**:
   $$
   r_1 r_2 = \sqrt{\frac{\alpha}{\beta}} \cdot \sqrt{\frac{\beta}{\alpha}} = 1
   $$

The equation with roots $$ r_1 $$ and $$ r_2 $$ is:
$$
x^2 + \frac{b}{\sqrt{ac}}x + 1 = 0
$$

### Additional Problem

If the roots of the equation $$ ax^2 + cx + c = 0 $$ are in the ratio $$ m:n $$, let the roots be $$ r_1 = km $$ and $$ r_2 = kn $$ for some $$ k $$.

Using Vieta's formulas:
- $$ r_1 + r_2 = km + kn = k(m+n) = -\frac{c}{a} $$
- $$ r_1 r_2 = km \cdot kn = k^2mn = \frac{c}{a} $$

From the first equation, we have $$ k = -\frac{c}{a(m+n)} $$.

Substituting $$ k $$ into the second equation:
$$
\left(-\frac{c}{a(m+n)}\right)^2 mn = \frac{c}{a}
$$
This leads to:
$$
\frac{c^2 mn}{a^2(m+n)^2} = \frac{c}{a}
$$
Cross-multiplying gives:
$$
c^2 mn = ac(m+n)^2
$$
This proves the relationship.

For the roots $$ (a\alpha + b)^{-1} $$ and $$ (a\beta + b)^{-1} $$, we can apply similar reasoning using Vieta's formulas to derive the new equation.
For the roots $$ \alpha \beta^{-1}, \beta \alpha^{-1} $$, the equation is $$ x^2 - \frac{b^2 - 2ac}{ac}x + 1 = 0 $$. For the roots $$ \alpha^2, \beta^2 $$, the equation is $$ x^2 - \frac{b^2 - 2ac}{a^2}x + \frac{c^2}{a^2} = 0 $$. For the roots $$ \alpha^3, \beta^3 $$, the equation is $$ x^2 + \frac{b(b^2 - 2ac)}{a^3}x + \frac{c^3}{a^3} = 0 $$. For the roots $$ \sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}} $$, the equation is $$ x^2 + \frac{b}{\sqrt{ac}}x + 1 = 0 $$. If the roots of $$ ax^2 + cx + c = 0 $$ are in the ratio $$ m:n $$, then the relationship is $$ c^2 mn = ac(m+n)^2 $$.

If and are the roots of , find the equations whose roots are

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