If $$ \alpha, \beta $$ are the roots of the equation $$ 3 x^{2}+5 x-1=0 $$ Construct a quadratic equation whose roots are (1) $$ 5 \alpha, 5 \beta $$ $$ \begin{array}{lll} ext { (2) } \alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha} & ext { (3) } \alpha^{3} ext { and } \beta^{3} ext { Assignment }\end{array} $$

Question

If $$ \alpha, \beta $$ are the roots of the equation $$ 3 x^{2}+5 x-1=0 $$ Construct a quadratic equation whose roots are (1) $$ 5 \alpha, 5 \beta $$ $$ \begin{array}{lll}	ext { (2) } \alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha} & 	ext { (3) } \alpha^{3} 	ext { and } \beta^{3} 	ext { Assignment }\end{array} $$

UpStudy AI · Accepted Answer

To construct the quadratic equations for the given roots, we will follow these steps:

1. **Find the roots $$ \alpha $$ and $$ \beta $$ of the original equation $$ 3x^2 + 5x - 1 = 0 $$**.
2. **Use the relationships between the roots and coefficients to find the new roots for each case**.
3. **Construct the quadratic equations based on the new roots**.

### Step 1: Find the roots $$ \alpha $$ and $$ \beta $$

The roots of the quadratic equation $$ ax^2 + bx + c = 0 $$ can be found using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For our equation $$ 3x^2 + 5x - 1 = 0 $$, we have $$ a = 3 $$, $$ b = 5 $$, and $$ c = -1 $$.

Let's calculate the roots $$ \alpha $$ and $$ \beta $$:

$$
\alpha, \beta = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}
$$

Calculating the discriminant:

$$
5^2 - 4 \cdot 3 \cdot (-1) = 25 + 12 = 37
$$

Now substituting back into the formula:

$$
\alpha, \beta = \frac{-5 \pm \sqrt{37}}{6}
$$

### Step 2: Construct the quadratic equations for the new roots

#### (1) Roots: $$ 5\alpha, 5\beta $$

The sum and product of the roots $$ 5\alpha $$ and $$ 5\beta $$ are:

$$
\text{Sum} = 5\alpha + 5\beta = 5(\alpha + \beta)
$$
$$
\text{Product} = 5\alpha \cdot 5\beta = 25\alpha\beta
$$

From Vieta's formulas for the original equation:

$$
\alpha + \beta = -\frac{b}{a} = -\frac{5}{3}
$$
$$
\alpha\beta = \frac{c}{a} = -\frac{1}{3}
$$

Now substituting these values:

$$
\text{Sum} = 5\left(-\frac{5}{3}\right) = -\frac{25}{3}
$$
$$
\text{Product} = 25\left(-\frac{1}{3}\right) = -\frac{25}{3}
$$

The quadratic equation with roots $$ 5\alpha $$ and $$ 5\beta $$ is:

$$
x^2 - \left(-\frac{25}{3}\right)x + \left(-\frac{25}{3}\right) = 0
$$

Multiplying through by 3 to eliminate the fraction:

$$
3x^2 + 25x + 25 = 0
$$

#### (2) Roots: $$ \alpha + \frac{1}{\beta}, \beta + \frac{1}{\alpha} $$

Calculating the sum and product of these roots:

$$
\text{Sum} = \left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) = (\alpha + \beta) + \left(\frac{1}{\beta} + \frac{1}{\alpha}\right)
$$

Using $$ \frac{1}{\beta} + \frac{1}{\alpha} = \frac{\alpha + \beta}{\alpha\beta} $$:

$$
\text{Sum} = -\frac{5}{3} + \frac{-\frac{5}{3}}{-\frac{1}{3}} = -\frac{5}{3} + 5 = \frac{10}{3}
$$

Now for the product:

$$
\text{Product} = \left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) = \alpha\beta + \alpha\cdot\frac{1}{\alpha} + \beta\cdot\frac{1}{\beta} + \frac{1}{\alpha\beta}
$$

This simplifies to:

$$
\text{Product} = -\frac{1}{3} + 1 + 1 - 3 = \frac{5}{3}
$$

The quadratic equation is:

$$
x^2 - \frac{10}{3}x + \frac{5}{3} = 0
$$

Multiplying through by 3:

$$
3x^2 - 10x + 5 = 0
$$

#### (3) Roots: $$ \alpha^3, \beta^3 $$

Using the identity $$ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) $$:

First, we find $$ \alpha^2 + \beta^2 $$:

$$
\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{5}{3}\right)^2 - 2\left(-\frac{1}{3}\right) = \frac{25}{9} + \frac{2}{3} = \frac{25}{9} + \frac{6}{9} = \frac{31}{9}
$$

Now substituting into the identity:

$$
\alpha^3 + \beta^3 = \left(-\frac{5}{3}\right)\left(\frac{31}{9} + \frac{1}{3}\right) = -\frac{5}{3}\left(\frac{31 + 3}{9}\right) = -\frac{5}{3}\left(\frac{34}{9}\right) = -\frac{170}{27}
$$

For the product $$ \alpha^3 \beta^3 = (\alpha\beta)^3 = \left(-\frac{1}{3}\right)^3 = -\frac{1}{27} $$.

The quadratic equation is:

$$
x^2 - \left(-\frac{170}{27}\right)x - \left(-\frac{1}{27}\right) = 0
$$

Multiplying through by 27:

$$
27x^2 + 170x + 1 = 0
$$

### Summary of Quadratic Equations

1. For roots $$ 5\alpha, 5\beta $$: $$ 3x^2 + 25x + 25 = 0 $$
2. For roots $$ \alpha + \frac{1}{\
1. Quadratic equation with roots $$5\alpha$$ and $$5\beta$$: $$3x^2 + 25x + 25 = 0$$ 2. Quadratic equation with roots $$\alpha + \frac{1}{\beta}$$ and $$\beta + \frac{1}{\alpha}$$: $$3x^2 - 10x + 5 = 0$$ 3. Quadratic equation with roots $$\alpha^3$$ and $$\beta^3$$: $$27x^2 + 170x + 1 = 0$$

If are the roots of the equation
Construct a quadratic equation whose roots are (1)

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