Answer :
Certainly! Let's solve this problem step-by-step:
### Step 1: Identify the roots of the given quadratic equation
The roots of the given quadratic equation [tex]\(3x^2 + 5x - 1 = 0\)[/tex] are:
[tex]\[ \alpha = -\frac{5}{6} + \frac{\sqrt{37}}{6}, \quad \beta = -\frac{\sqrt{37}}{6} - \frac{5}{6} \][/tex]
### Step 2: Construct equations for the given conditions
#### (a) Equation with roots [tex]\(5\alpha\)[/tex] and [tex]\(5\beta\)[/tex]
To construct the equation with roots [tex]\(5\alpha\)[/tex] and [tex]\(5\beta\)[/tex], we start by substituting [tex]\(x = 5t\)[/tex] into the original equation to form:
[tex]\[ 3\left(\frac{x}{5}\right)^2 + 5\left(\frac{x}{5}\right) - 1 = 0 \][/tex]
Simplifying, we get:
[tex]\[ \frac{3}{25}x^2 + \frac{5}{5}x - 1 = 0 \][/tex]
[tex]\[ \frac{3}{25}x^2 + x - 1 = 0 \][/tex]
Multiplying through by 25 to clear the fraction gives the final equation:
[tex]\[ 3x^2 + 25x - 25 = 0 \][/tex]
#### (b) Equation with roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]
To construct the equation with roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
[tex]\[ 3(t^{1/2})^2 + 5t^{1/2} - 1 = 0 \][/tex]
Where [tex]\(t = x^2\)[/tex], this simplifies to:
[tex]\[ 3x + 5\sqrt{x} - 1 = 0 \][/tex]
#### (c) Equation with roots [tex]\(\frac{1}{\alpha}\)[/tex] and [tex]\(\frac{1}{\beta}\)[/tex]
To construct the equation with the roots [tex]\(\frac{1}{\alpha}\)[/tex] and [tex]\(\frac{1}{\beta}\)[/tex]:
Change the original equation by substituting [tex]\(x = \frac{1}{t}\)[/tex]:
[tex]\[ 3\left(\frac{1}{x}\right)^2 + 5\left(\frac{1}{x}\right) - 1 = 0 \][/tex]
Multiply through by [tex]\(x^2\)[/tex] to clear the fractions:
[tex]\[ 3 + 5x - x^2 = 0 \][/tex]
Convert to:
[tex]\[ x^2 - 5x - 3 = 0 \][/tex]
#### (d) Equation with roots [tex]\(\alpha + \frac{1}{\beta}\)[/tex] and [tex]\(\beta + \frac{1}{\alpha}\)[/tex]
Roots are [tex]\(\alpha + \frac{1}{\beta}\)[/tex] and [tex]\(\beta + \frac{1}{\alpha}\)[/tex]. These roots are obtained by:
[tex]\[ \left(x - \left(\alpha + \frac{1}{\beta}\right)\right)\left(x - \left(\beta + \frac{1}{\alpha}\right)\right) = 0 \][/tex]
Using the specific values of [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ \left(x - \left(-\frac{5}{6} + \frac{\sqrt{37}}{6} + \frac{6}{1 -\frac{5}{6}} \right)\right) \left(x - \left(-\frac{\sqrt{37}}{6} - \frac{5}{6} \right.+ \frac{6}{1 + \frac{\sqrt{37}}{6} - \frac{5}{6}} \right)\right) = 0 \][/tex]
This simplifies to:
[tex]\[ 3x^2 - 10x + 4 = 0 \][/tex]
#### (e) Equation with roots [tex]\(\frac{\alpha+1}{β+1}\)[/tex] and [tex]\(\frac{\beta+1}{\alpha+1}\)[/tex]
To construct the equation with roots [tex]\(\frac{\alpha+1}{\beta+1}\)[/tex] and [tex]\(\frac{\beta+1}{\alpha+1}\)[/tex]:
[tex]\[ (x - \frac{-5/6 \ + \ \sqrt{37}/6 \ +1 \}{- \ (sqrt(37)\-6)})(X-(5)< \frac(sqrt(37)/6\right) ) = 0 \][/tex]
This simplifies to:
[tex]\[ 9x^2 + 19x + 9 = 0 \][/tex]
### Final compiled results
The equations based on the transformed roots are:
(a) [tex]\(3x^2 + 25x - 25 = 0\)[/tex]
(b) [tex]\(3x + 5\sqrt{x} - 1 = 0\)[/tex]
(c) [tex]\(x^2 - 5x - 3 = 0\)[/tex]
(d) [tex]\(3x^2 - 10x + 4 = 0\)[/tex]
(e) [tex]\(9x^2 + 19x + 9 = 0\)[/tex]
### Step 1: Identify the roots of the given quadratic equation
The roots of the given quadratic equation [tex]\(3x^2 + 5x - 1 = 0\)[/tex] are:
[tex]\[ \alpha = -\frac{5}{6} + \frac{\sqrt{37}}{6}, \quad \beta = -\frac{\sqrt{37}}{6} - \frac{5}{6} \][/tex]
### Step 2: Construct equations for the given conditions
#### (a) Equation with roots [tex]\(5\alpha\)[/tex] and [tex]\(5\beta\)[/tex]
To construct the equation with roots [tex]\(5\alpha\)[/tex] and [tex]\(5\beta\)[/tex], we start by substituting [tex]\(x = 5t\)[/tex] into the original equation to form:
[tex]\[ 3\left(\frac{x}{5}\right)^2 + 5\left(\frac{x}{5}\right) - 1 = 0 \][/tex]
Simplifying, we get:
[tex]\[ \frac{3}{25}x^2 + \frac{5}{5}x - 1 = 0 \][/tex]
[tex]\[ \frac{3}{25}x^2 + x - 1 = 0 \][/tex]
Multiplying through by 25 to clear the fraction gives the final equation:
[tex]\[ 3x^2 + 25x - 25 = 0 \][/tex]
#### (b) Equation with roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]
To construct the equation with roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
[tex]\[ 3(t^{1/2})^2 + 5t^{1/2} - 1 = 0 \][/tex]
Where [tex]\(t = x^2\)[/tex], this simplifies to:
[tex]\[ 3x + 5\sqrt{x} - 1 = 0 \][/tex]
#### (c) Equation with roots [tex]\(\frac{1}{\alpha}\)[/tex] and [tex]\(\frac{1}{\beta}\)[/tex]
To construct the equation with the roots [tex]\(\frac{1}{\alpha}\)[/tex] and [tex]\(\frac{1}{\beta}\)[/tex]:
Change the original equation by substituting [tex]\(x = \frac{1}{t}\)[/tex]:
[tex]\[ 3\left(\frac{1}{x}\right)^2 + 5\left(\frac{1}{x}\right) - 1 = 0 \][/tex]
Multiply through by [tex]\(x^2\)[/tex] to clear the fractions:
[tex]\[ 3 + 5x - x^2 = 0 \][/tex]
Convert to:
[tex]\[ x^2 - 5x - 3 = 0 \][/tex]
#### (d) Equation with roots [tex]\(\alpha + \frac{1}{\beta}\)[/tex] and [tex]\(\beta + \frac{1}{\alpha}\)[/tex]
Roots are [tex]\(\alpha + \frac{1}{\beta}\)[/tex] and [tex]\(\beta + \frac{1}{\alpha}\)[/tex]. These roots are obtained by:
[tex]\[ \left(x - \left(\alpha + \frac{1}{\beta}\right)\right)\left(x - \left(\beta + \frac{1}{\alpha}\right)\right) = 0 \][/tex]
Using the specific values of [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ \left(x - \left(-\frac{5}{6} + \frac{\sqrt{37}}{6} + \frac{6}{1 -\frac{5}{6}} \right)\right) \left(x - \left(-\frac{\sqrt{37}}{6} - \frac{5}{6} \right.+ \frac{6}{1 + \frac{\sqrt{37}}{6} - \frac{5}{6}} \right)\right) = 0 \][/tex]
This simplifies to:
[tex]\[ 3x^2 - 10x + 4 = 0 \][/tex]
#### (e) Equation with roots [tex]\(\frac{\alpha+1}{β+1}\)[/tex] and [tex]\(\frac{\beta+1}{\alpha+1}\)[/tex]
To construct the equation with roots [tex]\(\frac{\alpha+1}{\beta+1}\)[/tex] and [tex]\(\frac{\beta+1}{\alpha+1}\)[/tex]:
[tex]\[ (x - \frac{-5/6 \ + \ \sqrt{37}/6 \ +1 \}{- \ (sqrt(37)\-6)})(X-(5)< \frac(sqrt(37)/6\right) ) = 0 \][/tex]
This simplifies to:
[tex]\[ 9x^2 + 19x + 9 = 0 \][/tex]
### Final compiled results
The equations based on the transformed roots are:
(a) [tex]\(3x^2 + 25x - 25 = 0\)[/tex]
(b) [tex]\(3x + 5\sqrt{x} - 1 = 0\)[/tex]
(c) [tex]\(x^2 - 5x - 3 = 0\)[/tex]
(d) [tex]\(3x^2 - 10x + 4 = 0\)[/tex]
(e) [tex]\(9x^2 + 19x + 9 = 0\)[/tex]
