Chapter 3: Problem 218
If \(\alpha, \beta\) are roots of \(8 \mathrm{x}^{2}-3 \mathrm{x}+27=0\) then \(\left(\alpha^{2} / \beta\right)^{(1 / 3)}+\left(\beta^{2} / \alpha\right)^{(1 / 3)}=\) (a) \((1 / 3)\) (b) \((7 / 2)\) (c) 4 (d) \((1 / 4)\)
Short Answer
Expert verified
The sum \(\left(\alpha^{2} / \beta\right)^{1/3} + \left(\beta^{2} / \alpha\right)^{1/3}\) is an imaginary number, so none of the given options (a), (b), (c) or (d) is the correct answer for this exercise.
Step by step solution
01
Identify the Quadratic Equation and Roots
Firstly, we are given the quadratic equation \(8x^{2}-3x+27=0\) with roots α and β.
02
Use Vieta's Formulas for the Sum and Product of Roots
We know that for a quadratic equation \(ax^{2}+bx+c=0\), the sum of the roots is given by \(-b/a\) and the product of the roots by \(c/a\). So for our equation, the sum and product of the roots are:
$$\alpha + \beta = \frac{-(-3)}{8} = \frac{3}{8}$$
$$\alpha \beta = \frac{27}{8}$$
03
Multiply the given expression by its conjugate
We want to find the value of \(\left(\alpha^{2} / \beta\right)^{1/3} + \left(\beta^{2} / \alpha\right)^{1/3}\). To make the expression easier to work with, let's multiply the expression by the conjugate of the given expression:
$$\left(\left(\alpha^{2} / \beta\right)^{1/3} + \left(\beta^{2} / \alpha\right)^{1/3}\right)\left(\left(\alpha^{2} / \beta\right)^{1/3} - \left(\beta^{2} / \alpha\right)^{1/3}\right)$$
04
Simplify the expression
Now let's simplify the expression we got in step 3:
$$\left(\frac{\alpha^{2}}{\beta}\right) - \left(\frac{\beta^{2}}{\alpha}\right) = \frac{\alpha^{3}\beta - \alpha\beta^{3}}{\alpha\beta} = \frac{(\alpha^{3} - \beta^{3})}{\alpha\beta}$$
05
Use the sum and product of roots
Now, let's substitute the relationship we found in step 2 into the expression we got in step 4:
\[\frac{(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)}{\alpha\beta}
=\frac{\frac{3}{8}\left[(\alpha+\beta)^2-3\alpha\beta\right]}{\frac{27}{8}}
= \frac{3[(\frac{3}{8})^2 - \frac{9}{8}\frac{27}{8}]}{27}\]
06
Calculate the desired value
Now let's calculate the value of the expression above:
\[\frac{3\left[(\frac{9}{64}) - \frac{243}{64}\right]}{27}
= \frac{3(\frac{-234}{64})}{27}
= \frac{-234}{64} \cdot \frac{1}{9}
= -\frac{26}{8} = \frac{-13}{4}\]
07
Find the value of the initial expression
Since we multiplied our initial expression by its conjugate, we need to find the square root of the result we got in step 6 to get the answer for the initial expression. As the result is negative, the sum \(\left(\alpha^{2} / \beta\right)^{1/3} + \left(\beta^{2} / \alpha\right)^{1/3}\) is an imaginary number, which means none of the given options (a), (b), (c) or (d) is the correct answer for this exercise.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Roots
When tackling quadratic equations like the one in our exercise, understanding the concept of roots is crucial. Quadratic equations are typically expressed as \(ax^2 + bx + c = 0\), and they can have two roots. These roots are the values of \(x\) that satisfy the equation, essentially making it true. In our given quadratic equation, \(8x^2 - 3x + 27 = 0\), \(\alpha\) and \(\beta\) are these roots. Finding these roots usually involves factoring the quadratic, completing the square, or using the quadratic formula.
- For real and distinct roots: The discriminant \(b^2 - 4ac\) should be positive.
- For equal roots: The discriminant should be zero.
- For complex roots: The discriminant should be negative.
Vieta's Formulas
Vieta's formulas are powerful tools for dealing with polynomials, especially for problems involving the sum and product of roots. These formulas relate the coefficients of a polynomial to sums and products of its roots.For a quadratic equation \(ax^2 + bx + c = 0\), Vieta's formulas tell us:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a}\).
- The product of the roots \(\alpha \beta = \frac{c}{a}\).
Complex Numbers
Complex numbers can initially be intimidating, but they are simply an extension of the familiar number system used to solve problems where square roots of negative numbers appear. Introducing the imaginary unit \(i\), defined as \(i^2 = -1\), allows us to consider these extensions.A complex number is expressed as \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part. Complex numbers are crucial when dealing with the roots of polynomials, particularly when the discriminant of a quadratic equation is negative, as is likely in our exercise's scenario.
- A complex conjugate of a number is simply flipping the sign of the imaginary part, making complex roots often come in pairs like \(a + bi\) and \(a - bi\).
- Operations like addition or multiplication on complex numbers follow the usual rules, but remember the special case of \(i\).
Polynomials
Polynomials like the quadratic equation we dealt with are expressions consisting of variables, coefficients, and exponents arranged in a defined structure. In our exercise, the polynomial is a quadratic, meaning the highest exponent of the variable \(x\) is 2.A general polynomial is written as \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where \(a_n\) are coefficients and \(n\) indicates the degree of the polynomial.
- In quadratic terms, understanding the structure \(ax^2 + bx + c = 0\) provides insight into potential roots (solving one variable in terms of others).
- Each polynomial's degree determines the number of roots, including complex or repeated roots due to the fundamental theorem of algebra.