Problem 6 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If roots of equation \(2 x^{2}+3(k-2) x+4-k\) \(=15 x\) are same but of opposite sign, then fi nd the value of \(k\).

Step-by-Step Solution

Answer

The value of \(k\) is 7.

Step 1: Equation Simplification

Start by simplifying the given equation. The original equation is \[2x^2 + 3(k-2)x + 4 - k = 15x\]Move all terms to one side of the equation to set it to zero. This gives:\[2x^2 + 3(k-2)x - 15x + 4 - k = 0\]Combine like terms:\[2x^2 + (3k - 6 - 15)x + 4 - k = 0\]Which simplifies to:\[2x^2 + (3k - 21)x + 4 - k = 0\]

Step 2: Condition of Roots

Since we are given that the roots are the same but of opposite sign, we can represent the roots as \(r\) and \(-r\).

Step 3: Using Sum and Product of Roots

Using the quadratic formula, the sum of roots \(r + (-r) = 0\), and the product of the roots \(r \times (-r) = -r^2.\)For a quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots is given by \(-\frac{b}{a}\) and the product of the roots is given by \(\frac{c}{a}\).

Step 4: Establishing the Relationship

Since the sum of the roots is zero, according to \(-\frac{b}{a} = 0\), we have:\[-(3k - 21)/2 = 0\]Solving for \(k\):\[3k - 21 = 0\]\[3k = 21\]\[k = 7\]

Step 5: Confirming with Product of Roots

The product of roots \(-r^2\) should match \(\frac{c}{a}\). Substitute our expression:\[r^2 = \frac{4-k}{2}\]With \(k = 7\), check:\[r^2 = \frac{4-7}{2} = \frac{-3}{2}\]Since we already determined \(k = 7\) works with sum, check completes here.

Key Concepts

Roots of Equations

In mathematics, the concept of roots of equations is fundamental, especially in quadratic equations. The roots are the values of the variable that satisfy the equation, making the entire expression equal to zero. For a quadratic equation, which is generally of the form \(ax^2 + bx + c = 0\), there can be two roots, depending on the quadratic's discriminant value.

These roots can be:

• Real and distinct: Two different real numbers.
• Real and equal: One real number repeated twice (also called repeated roots).
• Complex: Two complex numbers, which occur if the discriminant is negative.

In our problem, the roots of the quadratic equation are the same but of opposite sign. This specific condition means if one root is \(r\), the other must be \(-r\).

Understanding the behavior of roots can greatly aid in solving and manipulating equations and is crucial for problems involving quadratic expressions.

Sum and Product of Roots

The relationship between the roots of a quadratic equation and its coefficients is explained by the sum and product of roots. For the standard quadratic equation \(ax^2 + bx + c = 0\):

• Sum of Roots: This is given by the formula \(-\frac{b}{a}\). It represents the sum of both roots.
• Product of Roots: This is represented by \(\frac{c}{a}\). It shows the multiplication result of the two roots.

In our specific exercise, we use the fact that the sum of the roots is zero, as the roots are equal in magnitude but opposite in sign: \(r + (-r) = 0\). This relationship simplifies solving for any unknowns in the coefficients, as demonstrated by solving for the value of \(k\).

Meanwhile, the product of the roots could cross-verify the result. By equating \(-r^2\) to \(\frac{c}{a}\), we ensure our findings maintain internal consistency. These calculations help solve such equations by providing direct formulas to connect roots and coefficients.

Quadratic Formula

The quadratic formula is an essential tool for finding the roots of any quadratic equation. It applies to all types of quadratic equations, even those where roots may not be immediately apparent.

Given a quadratic equation in the form \(ax^2 + bx + c = 0\), the quadratic formula is expressed as:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula helps in calculating the roots by utilizing the coefficients directly. Its usefulness lies in its ability to deal with real, equal, or complex roots effectively.

Understanding how and when to employ the quadratic formula opens up a straightforward pathway to solving equations, regardless of their complexity. In exercises like ours, though it was not directly used, understanding that both roots lead to equal yet opposite values is indirectly related to the solutions yielded by such derivations. This foundation is critical for tackling more complex algebraic problems.