Problem 82 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find \(k\) if the equations \(4 x^{2}-11 x+2 k=0\) and \(x^{2}-3 x-k=0\) have a common root and obtain the common root for this value of \(k\). \\{Ans. \(k=0\) or \(-\frac{17}{36}\), common root 0 or \(\left.\frac{17}{6}\right\\}\)

Step-by-Step Solution

Answer

First, let's consider the case where \(r_1 = 0\). In this case, we have: 1) \(0 + r_2 = 3\), which implies \(r_2 = 3\) 2) \(0 \cdot r_2 = -k\), which implies \(k = 0\) So for one solution, we have \(k = 0\), and the common root is \(r = 0\). Now let's consider the case where \(r_1 \neq 0\). In this case, since \(r_1 \cdot r_2 = -k\), we have: \[ k = -\frac{1}{6}r_1 \cdot r_2 \] We know that \(r_1 + r_2 = 3\), so we can express \(r_2\) in terms of \(r_1\): \[ r_2 = 3 - r_1 \] Substitute this expression for \(r_2\) into the equation for k: \[ k = -\frac{1}{6}r_1(3 - r_1) \] Now, we can solve for \(r_1\): 1) Simplify the equation: \(6k = -r_1(3 - r_1)\) 2) Expand the equation: \(6k = -3r_1 + r_1^2\) 3) Rearrange the equation: \(r_1^2 - 3r_1 - 6k = 0\) With the equation for the common root, we can now substitute the value of \(k = -\frac{17}{36}\) to find the common root: \[ r_1^2 - 3r_1 - 6 \times -\frac{17}{36} = 0 \] \[ r_1^2 - 3r_1 + \frac{17}{6} = 0 \] Solving for \(r_1\), we find that the common root is \(r = \frac{17}{6}\). Therefore, the possible values of \(k\) are \(k = 0\) and \(k = -\frac{17}{36}\), with the corresponding common roots \(r = 0\) and \(r = \frac{17}{6}\).

Step 1: Express x in terms of k from the first equation

Divide the first equation by 4 to simplify it: \[ x^2 - \frac{11}{4}x + \frac{k}{2} = 0 \] Let the common root be \(x = r\). We know that the sum and product of the roots of a quadratic equation \(ax^2 + bx + c = 0\) are given by \(-\frac{b}{a}\) and \(\frac{c}{a}\), respectively. So, we have: \[ r^2 - \frac{11}{4}r + \frac{k}{2} = 0 \]

Step 2: Substitute the expression of r in the second equation

Plug the value of \(x = r\) into the second equation: \[ r^2 - 3r - k = 0 \]

Step 3: Solve for k

Now we compare the two equations obtained in Steps 1 and 2: \[ r^2 - \frac{11}{4}r + \frac{k}{2} = r^2 - 3r - k \] To solve for k, subtract the second equation from the first equation: \[ -\frac{11}{4}r + \frac{k}{2} + 3r + k = 0 \] Simplifying, we get: \[ \frac{1}{4}r = \frac{3}{2}k \] Divide both sides by \(\frac{3}{2}\) to find the value of k: \[ k = \frac{1}{6}r \] Since we know that there are only two possible values for the common root, we can find them by considering the two roots of the second equation. Let the two roots of the second equation be \(r_1\) and \(r_2\). Then: 1) \(r_1 + r_2 = 3\) 2) \(r_1 \cdot r_2 = -k\) We can find the value of \(k\) using these two equations.

Key Concepts

Common Roots

When given two quadratic equations, finding a common root allows us to understand how these equations share at least one solution. In mathematical terms, a "root" is a value of the variable that satisfies the equation, meaning it makes the equation true when plugged in. If two distinct quadratic equations share a root, it implies both equations have at least one solution in common.

• A common root means plugging the value into both equations will yield zero for each.
• In our example, we are checking which values of \( x \) satisfy both equations \( 4x^2 - 11x + 2k = 0 \) and \( x^2 - 3x - k = 0 \).

Determining a common root involves solving a system of equations, which might require techniques such as substitution or elimination. Here, we substitute the potential root \( r \) into both equations, simplifying and comparing them to find the possible values of \( k \) and \( r \). This approach solutions for \( k \) by setting expressions for \( r \) from both equations equal to each other.

Sum and Product of Roots

The sum and product of roots are essential properties that help in understanding and solving quadratic equations. For a quadratic equation in the form \( ax^2 + bx + c = 0 \), the sum and product of its roots \( r_1 \) and \( r_2 \) can be calculated using coefficients:

• The sum of the roots: \( r_1 + r_2 = -\frac{b}{a} \)
• The product of the roots: \( r_1 \cdot r_2 = \frac{c}{a} \)

In the context of the exercise, we use these relationships as tools for developing equations based on given roots. With one known equation \( x^2 - 3x - k = 0 \), the sum and product of its roots become:

• \( r_1 + r_2 = 3 \)
• \( r_1 \cdot r_2 = -k \)

These relationships help create systems of equations that can be solved to identify the common root \( r \) and values of \( k \), aiding in the primary task of finding shared roots or understanding their interactions across different equations.

Quadratic Formula

The quadratic formula is a powerful mathematical tool used to find the roots of any quadratic equation given in the standard form \( ax^2 + bx + c = 0 \). The formula itself is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula always provides a solution to a quadratic equation, and it's especially useful when factoring is challenging or not possible. Using the quadratic formula involves the following steps:

• Identify the coefficients \( a \), \( b \), and \( c \) from the given equation.
• Calculate the discriminant \( b^2 - 4ac \).
• Plug these values into the formula to solve for \( x \).

In our exercise, while the main focus was on common roots, the quadratic formula can be a helpful alternative method for verifying the values of \( x \) known from other algebraic manipulations. By calculating the discriminant and applying the quadratic root solutions, we can double-check our possible solutions for \( k \) and ensure accuracy in identifying common roots.