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If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.
(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube
(b)the equation whose roots are alpha cube and beta cube?
(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube
(b)the equation whose roots are alpha cube and beta cube?
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If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) ...
Finding the Roots of the Quadratic Equation
To find the roots (α and β) of the equation \(2x^2 - 5x - 1 = 0\), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 2\), \(b = -5\), and \(c = -1\).
- Discriminant:
\[
D = b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot (-1) = 25 + 8 = 33
\]
- Roots:
\[
\alpha, \beta = \frac{5 \pm \sqrt{33}}{4}
\]
Finding α³ and β³
To find \(α^3\) and \(β^3\), we can use the identity:
\[
α^3 + β^3 = (α + β)(α^2 - αβ + β^2)
\]
Using Vieta's formulas:
- \(α + β = \frac{5}{2}\)
- \(αβ = -\frac{1}{2}\)
Calculate \(α^2 + β^2\):
\[
α^2 + β^2 = (α + β)^2 - 2αβ = \left(\frac{5}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{25}{4} + 1 = \frac{29}{4}
\]
Now, substituting into the identity:
\[
α^3 + β^3 = \left(\frac{5}{2}\right) \left(\frac{29}{4} + \frac{1}{2}\right) = \left(\frac{5}{2}\right) \left(\frac{29 + 2}{4}\right) = \left(\frac{5}{2}\right) \left(\frac{31}{4}\right) = \frac{155}{8}
\]
Finding \( \frac{1}{α^3} + \frac{1}{β^3} \)
Using the relation:
\[
\frac{1}{α^3} + \frac{1}{β^3} = \frac{α^3 + β^3}{α^3β^3}
\]
We know \(α^3β^3 = (αβ)^3 = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}\).
Thus:
\[
\frac{1}{α^3} + \frac{1}{β^3} = \frac{\frac{155}{8}}{-\frac{1}{8}} = -155
\]
Finding the Equation with Roots α³ and β³
The required equation can be formed using the roots α³ and β³:
1. Sum of roots:
\[
α^3 + β^3 = \frac{155}{8}
\]
2. Product of roots:
\[
α^3β^3 = -
To find the roots (α and β) of the equation \(2x^2 - 5x - 1 = 0\), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 2\), \(b = -5\), and \(c = -1\).
- Discriminant:
\[
D = b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot (-1) = 25 + 8 = 33
\]
- Roots:
\[
\alpha, \beta = \frac{5 \pm \sqrt{33}}{4}
\]
Finding α³ and β³
To find \(α^3\) and \(β^3\), we can use the identity:
\[
α^3 + β^3 = (α + β)(α^2 - αβ + β^2)
\]
Using Vieta's formulas:
- \(α + β = \frac{5}{2}\)
- \(αβ = -\frac{1}{2}\)
Calculate \(α^2 + β^2\):
\[
α^2 + β^2 = (α + β)^2 - 2αβ = \left(\frac{5}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{25}{4} + 1 = \frac{29}{4}
\]
Now, substituting into the identity:
\[
α^3 + β^3 = \left(\frac{5}{2}\right) \left(\frac{29}{4} + \frac{1}{2}\right) = \left(\frac{5}{2}\right) \left(\frac{29 + 2}{4}\right) = \left(\frac{5}{2}\right) \left(\frac{31}{4}\right) = \frac{155}{8}
\]
Finding \( \frac{1}{α^3} + \frac{1}{β^3} \)
Using the relation:
\[
\frac{1}{α^3} + \frac{1}{β^3} = \frac{α^3 + β^3}{α^3β^3}
\]
We know \(α^3β^3 = (αβ)^3 = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}\).
Thus:
\[
\frac{1}{α^3} + \frac{1}{β^3} = \frac{\frac{155}{8}}{-\frac{1}{8}} = -155
\]
Finding the Equation with Roots α³ and β³
The required equation can be formed using the roots α³ and β³:
1. Sum of roots:
\[
α^3 + β^3 = \frac{155}{8}
\]
2. Product of roots:
\[
α^3β^3 = -
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If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube(b)the equation whose roots are alpha cube and beta cube?
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If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube(b)the equation whose roots are alpha cube and beta cube? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube(b)the equation whose roots are alpha cube and beta cube? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube(b)the equation whose roots are alpha cube and beta cube?.
If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube(b)the equation whose roots are alpha cube and beta cube? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube(b)the equation whose roots are alpha cube and beta cube? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If alpha and beta are the roots of the equation; 2x^2-5x-1=0,find.(a) the values of alpha cube and beta cube and 1/alpha cube + beta cube(b)the equation whose roots are alpha cube and beta cube?.
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