Problem 31 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If the equation \(\left(k^{2}-5 k+6\right) x^{2}+\left(k^{2}-3 k+2\right) x+\left(k^{2}-4\right)=0\) is satisfied by more than two values of \(x\), then determine the value of \(k\).

Step-by-Step Solution

Answer

The value of \(k\) is 2.

Step 1: Setting coefficients to zero

First set the coefficients of the quadratic equation (the expressions in terms of \(k\)) equal to zero:1) \(k^2 - 5k + 6 = 0\)2) \(k^2 - 3k + 2 = 0\)3) \(k^2 - 4 = 0\)

Step 2: Solving equations

Next, solve these equations for \(k\):For 1) \(k^2 - 5k + 6 = 0\)The roots are \(k_1=2\) and \(k_2=3\).For 2) \(k^2 - 3k + 2 = 0\)The roots are \(k_1=1\) and \(k_2=2\).For 3) \(k^2 - 4 = 0\)The roots are \(k_1= -2\) and \(k_2=2\).

Step 3: Identify the common root

Now we need to identify the roots that are common across all three equations. In this case, \(k=2\) is common in all three equations.

Key Concepts

Solving Quadratic Equations

Quadratic equations are fundamental in algebra, and learning to solve them is a critical milestone for students. A quadratic equation is typically in the form of \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a \) is not zero. To solve such equations, you can use various methods such as factoring, using the quadratic formula, completing the square, or graphing.

For example, considering the quadratic equation \( k^2 - 5k + 6 = 0 \), you could factor it into \( (k - 2)(k - 3) = 0 \). This reveals the roots of the equation, since for two numbers to multiply to zero, one or both must be zero; thus, \( k - 2 = 0 \) or \( k - 3 = 0 \), giving the roots \( k = 2 \) or \( k = 3 \). When solving quadratic equations, it's essential to remember that there can be either one or two real roots, or none if the roots are complex.

Roots of Equations

The roots of an equation are the solutions that satisfy the equation, meaning they are the values of the variable that make the equation true. In the context of quadratic equations, roots are also called zeros. They can be real or complex, and a quadratic equation has at most two real roots. These roots can be distinct, meaning they have different values, or they may be equal, resulting in one solution, which is called a repeated root.

Using the exercise's equations as examples, we can find the roots by setting each quadratic equation to zero and solving for \(k\). For instance, the equation \( k^2 - 3k + 2 = 0 \) can be factored to give \( (k - 1)(k - 2) = 0 \), leading to the roots \( k = 1 \) and \( k = 2 \). Understanding the concept of roots is pivotal because it not only applies to quadratic equations but also to various forms of higher-degree equations.

Quadratic Equation Common Roots

Sometimes, we come across multiple quadratic equations that are expected to have common roots. Common roots are roots that are solutions to more than one quadratic equation. In the exercise, we have three different forms of quadratic equations, each associated with a different value of \(k\), and our goal is to find the value of \(k\) common to all.

The equations given in the exercise, when solved individually, provide different roots for the variable \(k\). Upon comparing the roots, \(k = 2\) stands out as a common root among all the equations. This concept of common roots is significant because it often comes up in systems of equations, where finding a common solution that satisfies all equations in the system is necessary. Recognizing shared roots helps in solving more complex algebraic problems, where multiple conditions must be met simultaneously.