Problem 15
Question
Let \(a, b, c\) be real numbers with \(a \neq 0\) and let \(\alpha, \beta\) be the roots of the equation \(a x^{2}+b x+c=0 .\) Express the roots of \(a^{3} x^{2}+a b c x+c^{3}=0\) in terms of \(\alpha, \beta\).
Step-by-Step Solution
Verified Answer
The roots of the equation \(a^3x^2 + abc x + c^3=0\) can be expressed in terms of the roots of the equation \(ax^2 + bx + c=0\) as:
\(\gamma = \alpha\)
\(\delta = \beta\)
1Step 1: Identify the roots of the given equations
Given the equation \(ax^2 + bx + c = 0\) has roots \(\alpha\) and \(\beta\). Also, we need to find the roots of the equation \(a^3x^2 + abc x + c^3 = 0\), let's denote these roots as \(\gamma\) and \(\delta\).
2Step 2: Utilize Vieta's formulas
For the equation \(ax^2 + bx + c = 0\), we have the following relationships:
Sum of roots:
\( \alpha + \beta = -\frac{b}{a} \) (1)
Product of roots:
\( \alpha \beta = \frac{c}{a} \) (2)
For the equation \(a^3x^2 + abc x + c^3 = 0\), we have the following relationships:
Sum of roots:
\( \gamma + \delta = -\frac{abc}{a^3} = -\frac{b}{a} \) (3)
Product of roots:
\( \gamma \delta = \frac{c^3}{a^3} = \frac{c}{a} \) (4)
3Step 3: Relate the roots of the two equations
From (1) and (3), we get:
\( \alpha + \beta = \gamma + \delta \)
From (2) and (4), we get:
\( \alpha \beta = \gamma \delta \)
Comparing these results, we can conclude the following relationship for the roots of the two quadratic equations:
The roots of the equation \(a^3x^2 + abc x + c^3=0\) can be expressed in terms of the roots of the equation \(ax^2 + bx + c=0\) as:
\(\gamma = \alpha\)
\(\delta = \beta\)
Key Concepts
Vieta's FormulasRoots of Quadratic EquationsPolynomial Relationships
Vieta's Formulas
Vieta's formulas provide a beautiful connection between the coefficients of a polynomial equation and its roots. Specifically, for a quadratic equation of the form \( ax^2 + bx + c = 0 \), Vieta's formulas state:
- The sum of the roots, \( \alpha + \beta \), is given by \( -\frac{b}{a} \).
- The product of the roots, \( \alpha \beta \), is given by \( \frac{c}{a} \).
Roots of Quadratic Equations
The roots of quadratic equations hold essential clues about its graph, and they are the points where the quadratic crosses the x-axis. Consider the general form of a quadratic equation: \( ax^2 + bx + c = 0 \). Solving for the roots involves finding distinct points \( x \) such that the equation equals zero. This is often done using the quadratic formula:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Whether the parabola intersects the x-axis.
- The nature of the roots: real and distinct, real and repeated, or complex, depending on the discriminant \( b^2 - 4ac \).
Polynomial Relationships
Polynomial relationships are essential for understanding the connections between different polynomial equations and their roots. In this exercise, you are asked to find roots of the polynomial \( a^3x^2 + abc x + c^3 = 0 \) in terms of roots \( \alpha \) and \( \beta \) of another equation. By applying Vieta’s formulas, we derive connections between coefficients and roots.Understanding that polynomial equations of the same degree can share the same roots if their coefficients relate in a particular way is foundational:
- The sum of the roots \( \gamma + \delta \) matches with \( \alpha + \beta \) based on the coefficients.
- The product of the roots \( \gamma \delta \) also reflects the coefficients’ relationship just like \( \alpha \beta \).
Other exercises in this chapter
Problem 13
Two candidates attempt to solve a quadratic of the form \(x^{2}+p x+q=0 .\) One starts with a wrong value of \(p\) and finds the roots to be 2 and 6 . The other
View solution Problem 14
If \(\alpha\) be a root of \(4 x^{2}+2 x-1=0\), prove that \(4 \alpha^{3}-3 \alpha\) is the other root.
View solution Problem 16
If \(\alpha+\beta=3\) and \(\alpha^{3}+\beta^{3}=7\), then show that \(\alpha\) and \(\beta\) are the roots of \(9 x^{2}-27 x+20=0\).
View solution Problem 17
If \(\alpha, \beta\) be the roots of \(a x^{2}+2 b x+c=0\) and \(\alpha+\delta, \beta+\delta\) be those of \(A x^{2}+2 B x+C=0\), then prove that \(\frac{b^{2}-
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