Problem 51
Question
If \(\alpha\) and \(\beta\) are the roots of \(a x^{2}+b x+c=0\), find the equation whose roots are given below:- i. \(\quad \frac{1}{\alpha+\beta}, \frac{1}{\alpha}+\frac{1}{\beta} .\left\\{\right.\) ii. \(\frac{\alpha}{\beta}, \frac{\beta}{\alpha} .\left\\{\right.\) iii. \(\alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha} .\left\\{\right.\) iv. \(\alpha^{2}+\beta^{2}, \frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}} .\)c\right)\left(a^{2}+c^{2}\right) x+\left(b^{2}-2 a c\right)^{2}=0\right\\}\( v. \)\frac{1}{a \alpha+b}, \frac{1}{a \beta+b}$.
Step-by-Step Solution
Verified Answer
The solutions for the given parts are as follows: i. \(x^2 + \frac{ab^2 - a^3}{bc}x + \frac{-a^3}{b^2} = 0\), ii. \(x^2 - (a + c)x + 1 = 0\), iii. \(x^2 - (\frac{c^2-a^2}{a})x + 1 = 0\), iv. \(x^2 - \frac{(b^4+a^4-2c^2a^2)}{a^2c^2}x + 4(c^2-a^2) = 0.\), and v. \(x^2 +( \frac{-2^a c^2}{b^2-a^2 c^2 })x + \frac{1}{b^2 - a^2 c^2} = 0\).
1Step 1: Solve for the roots in Part i
Given roots are \(\frac{1}{\alpha+\beta}\) and \(\frac{1}{\alpha}+\frac{1}{\beta}\). Expressing these roots using \(-\frac{b}{a}\) and \(\frac{c}{a}\), the roots become \(\frac{-a}{b}\) and \(\frac{a^2}{c}\). Hence, the quadratic equation is \(x^2 - (\frac{-a}{b} + \frac{a^2}{c})x + \frac{-a}{b} \times \frac{a^2}{c} = 0\), which simplifies to \(x^2 + \frac{ab^2 - a^3}{bc}x + \frac{-a^3}{b^2} = 0\).
2Step 2: Solve for the roots in Part ii
Given roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\). Expressing these roots using \(-\frac{b}{a}\) and \(\frac{c}{a}\), the roots become \(\frac{a}{c}\) and \(\frac{c}{a}\). Hence, the quadratic equation is \(x^2 - (\frac{a}{c} + \frac{c}{a})x + \frac{a}{c} \times \frac{c}{a} = 0\), which simplifies to \(x^2 - (a + c)x + 1 = 0\).
3Step 3: Solve for the roots in Part iii
Given roots are \(\alpha+\frac{1}{\beta}\) and \(\beta+\frac{1}{\alpha}\). Expressing these roots using \(-\frac{b}{a}\) and \(\frac{c}{a}\), the roots become \(-\frac{b}{a}+ \frac{a}{c}\) and \(-\frac{b}{a} + \frac{c}{a}\). Hence, the quadratic equation is \(x^2 - ((-\frac{b}{a}+ \frac{a}{c}) + (-\frac{b}{a} + \frac{c}{a}))x + ((-\frac{b}{a}+ \frac{a}{c}) \times (-\frac{b}{a} + \frac{c}{a})) = 0\), which simplifies to \(x^2 - (\frac{c^2-a^2}{a})x + 1 = 0\).
4Step 4: Solve for the roots in Part iv
Given roots are \(\alpha^{2}+\beta^{2}\) and \(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\). Expressing these roots using \(-\frac{b}{a}\) and \(\frac{c}{a}\), the roots become \(\frac{b^2}{a^2}-2c\) and \(\frac{a^2}{c^2}-2\). Hence, the quadratic equation is \(x^2 - ((\frac{b^2}{a^2}-2c)+(\frac{a^2}{c^2}-2))x + ((\frac{b^2}{a^2}-2c)\times(\frac{a^2}{c^2}-2))= 0\), that simplifies to \(x^2 - \frac{(b^4+a^4-2c^2a^2)}{a^2c^2}x + 4(c^2-a^2) = 0.\)
5Step 5: Solve for the roots in Part v
Given roots are \(\frac{1}{a \alpha+b}\) and \(\frac{1}{a \beta+b}\). Expressing these roots using \(-\frac{b}{a}\) and \(\frac{c}{a}\), the roots become \(\frac{1}{-b+ac}\) and \(\frac{1}{-b-ac}\). Hence, the quadratic equation is \(x^2 - ((\frac{1}{-b+ac}) + (\frac{1}{-b-ac}))x + ((\frac{1}{-b+ac}) (\frac{1}{-b-ac})) = 0\), which simplifies to \(x^2 +( \frac{-2^a c^2}{b^2-a^2 c^2 })x + \frac{1}{b^2 - a^2 c^2} = 0\).
Key Concepts
Roots of a Quadratic EquationSum and Product of RootsAlgebraic Transformations
Roots of a Quadratic Equation
Quadratic equations are fundamental algebraic expressions that have the general form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). An important aspect of these equations is discovering the values of \(x\) that make the equation true, known as 'roots' or 'solutions'. These roots can be found using various methods such as factoring, completing the square, graphing, or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
In the exercise provided, the problem revolves around manipulating the roots of one quadratic equation to find another equation with different roots. This not only tests the algebraic manipulation skills of a student but also deepens the understanding of the relationships between coefficients and roots of a quadratic equation. Determining these new roots involves operations like addition, multiplication, inversion, and combining these transformations to express them in terms of the original roots \(\alpha\) and \(\beta\).
The approach highlighted in the steps involves expressing the desired roots in terms of \(-\frac{b}{a}\) and \(\frac{c}{a}\), which correspond to the sum and product of the roots of the original quadratic equation, respectively. This method is a practical illustration of the general principles concerning the roots of a quadratic equation, underscoring the importance of mastery in this area for students preparing for competitive exams like IIT JEE.
In the exercise provided, the problem revolves around manipulating the roots of one quadratic equation to find another equation with different roots. This not only tests the algebraic manipulation skills of a student but also deepens the understanding of the relationships between coefficients and roots of a quadratic equation. Determining these new roots involves operations like addition, multiplication, inversion, and combining these transformations to express them in terms of the original roots \(\alpha\) and \(\beta\).
The approach highlighted in the steps involves expressing the desired roots in terms of \(-\frac{b}{a}\) and \(\frac{c}{a}\), which correspond to the sum and product of the roots of the original quadratic equation, respectively. This method is a practical illustration of the general principles concerning the roots of a quadratic equation, underscoring the importance of mastery in this area for students preparing for competitive exams like IIT JEE.
Sum and Product of Roots
One of the elegant connections in algebra is the relationship between the coefficents of a quadratic equation and its roots. According to Vieta's formulas, for a quadratic equation \(ax^2 + bx + c = 0\), the sum of its roots \(\alpha\) and \(\beta\) is equal to \(-\frac{b}{a}\) and the product of its roots is equal to \(\frac{c}{a}\).
This concept is crucial when solving for the new quadratic equation in each part of the exercise. For instance, when we need to find an equation whose roots are inversely related to the sum and product of the original roots, such as \(\frac{1}{\alpha+\beta}\) and \(\frac{1}{\alpha} + \frac{1}{\beta}\), we apply the relationships of sum and product directly to derive a new equation.
In practice, to improve understanding of this concept, one might consider manipulating common algebraic expressions to gain a better intuition about how these relationships work. For example, if students explore these relationships by substituting actual numbers for \(\alpha\) and \(\beta\), they can see firsthand how the coefficients change as a result. This hands-on strategy solidifies the theory behind sum and product of roots, making it a powerful tool in algebraic problem-solving, especially in competitive exam settings.
This concept is crucial when solving for the new quadratic equation in each part of the exercise. For instance, when we need to find an equation whose roots are inversely related to the sum and product of the original roots, such as \(\frac{1}{\alpha+\beta}\) and \(\frac{1}{\alpha} + \frac{1}{\beta}\), we apply the relationships of sum and product directly to derive a new equation.
In practice, to improve understanding of this concept, one might consider manipulating common algebraic expressions to gain a better intuition about how these relationships work. For example, if students explore these relationships by substituting actual numbers for \(\alpha\) and \(\beta\), they can see firsthand how the coefficients change as a result. This hands-on strategy solidifies the theory behind sum and product of roots, making it a powerful tool in algebraic problem-solving, especially in competitive exam settings.
Algebraic Transformations
Algebraic transformations involve reshaping algebraic expressions in various ways while maintaining their equality. These techniques are pivotal in deriving a new quadratic equation from the roots of an existing one. Transformations can include basic operations such as addition, subtraction, multiplication, division, and even functional operations like inversion and exponentiation.
In the provided solutions, a series of transformations allow us to express the desired new roots in relation to the original roots \(\alpha\) and \(\beta\), ultimately leading to a new quadratic equation. For instance, in part iv of the exercise, we are dealing with roots that are transformations of \(\alpha^2 + \beta^2\), requiring algebraic manipulation to extract a meaningful equation.
By practicing these algebraic transformations, students enhance their problem-solving flexibility and adaptability. A good tip is to always verify the transformed expressions by using the original equation's roots. With repeated exposure to different types of transformations and a methodical approach, students become better prepared to tackle the varied and complex algebraic problems they will encounter in exams like the IIT JEE.
In the provided solutions, a series of transformations allow us to express the desired new roots in relation to the original roots \(\alpha\) and \(\beta\), ultimately leading to a new quadratic equation. For instance, in part iv of the exercise, we are dealing with roots that are transformations of \(\alpha^2 + \beta^2\), requiring algebraic manipulation to extract a meaningful equation.
By practicing these algebraic transformations, students enhance their problem-solving flexibility and adaptability. A good tip is to always verify the transformed expressions by using the original equation's roots. With repeated exposure to different types of transformations and a methodical approach, students become better prepared to tackle the varied and complex algebraic problems they will encounter in exams like the IIT JEE.
Other exercises in this chapter
Problem 49
Find the equation whose roots are \((\alpha+\beta)^{2}\) and \((\alpha-\beta)^{2}\), where \(\alpha, \beta\) are the roots of \(2 x^{2}+2(m+n) x+\left(m^{2}+n^{
View solution Problem 50
If \(\alpha\) and \(\beta\) are the roots of \(x^{2}-2 x+3=0\), find the equation whose roots are:i. \(\quad \alpha+2, \beta+2 .\left\\{\right.\) ii. \(\frac{\a
View solution Problem 52
If \(\alpha, \beta\) are the roots of \(a x^{2}+b x+c=0, \alpha_{1},-\beta\) are the roots of \(a_{1} x^{2}+b_{1} x+c_{1}=0\), show that \(\alpha, \alpha_{1}\)
View solution Problem 53
Form an equation whose roots are negative of the roots of the equation \(x^{3}-5 x^{2}-7 x-3=0\).
View solution