Get the answers you've been searching for with IDNLearn.com. Get step-by-step guidance for all your technical questions from our dedicated community members.
Sagot :
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]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.