Problem 56 - A Complete Resource Book in Mathematics for JEE Main

Question

If the roots of the equation \(x^{2}-2 a x+a^{2}+a-3=0\) are real and less than 3 , then (A) \(a<2\) (B) \(2 \leq a \leq 3\) (B) \(3 \leq a \leq 4\) (D) \(a>4\)

Step-by-Step Solution

Answer

The answer is option (B): \(2 \leq a \leq 3\).

Step 1: Understand the Conditions for Real Roots

To ensure that the roots are real, the discriminant of the quadratic equation must be non-negative. The given equation is \(x^2 - 2ax + (a^2 + a - 3) = 0\). The standard quadratic formula \(ax^2 + bx + c = 0\) shows the discriminant is \(b^2 - 4ac\). For real roots, we need to calculate \((2a)^2 - 4(1)(a^2 + a - 3)\) and ensure it is \( \geq 0\).

Step 2: Compute the Discriminant

The discriminant for the equation is \(b^2 - 4ac\) where \(b = -2a\), \(a = 1\), and \(c = a^2 + a - 3\). Hence, the discriminant is: \((2a)^2 - 4(1)(a^2 + a - 3) = 4a^2 - 4(a^2 + a - 3)\). Simplifying, it becomes \(4a^2 - 4a^2 - 4a + 12 = -4a + 12\). For the roots to be real, \(-4a + 12 \geq 0\).

Step 3: Solve the Inequality for the Discriminant

Solve \(-4a + 12 \geq 0\):\[-4a + 12 \geq 0\12 \geq 4a\3 \geq a\]Therefore, \(a \leq 3\).

Step 4: Analyze the Roots Being Less than 3

We require the roots of the equation to be less than 3. Substitute \(x = 3\) into the inequality derived from the equation: \[3^2 - 2a(3) + a^2 + a - 3 < 0\] Simplify: \[9 - 6a + a^2 + a - 3 < 0\a^2 - 5a + 6 < 0\] Factor the quadratic inequality: \((a-2)(a-3) < 0\). This shows that \(2 < a < 3\).

Step 5: Combine the Conditions

Combining the results from both conditions: The discriminant condition \(a \leq 3\) and the root condition \(2 < a < 3\), we deduce that \(2 < a \leq 3\). Hence, the answer is option (B): \(2 \leq a \leq 3\).

Key Concepts

Discriminant: The Key to Real Roots

In quadratic equations, the discriminant is a crucial component that determines the nature of the roots. For any quadratic equation in the standard form \(ax^2 + bx + c = 0\), the discriminant \(D\) is given by the formula \(D = b^2 - 4ac\). The discriminant reveals whether the roots are real or complex and how they are spaced.

• If \(D > 0\), the equation has two distinct real roots.
• If \(D = 0\), there is exactly one real root, also known as a repeated or double root.
• If \(D < 0\), the equation has two complex roots, which are not real.

In the exercise provided, to ensure the roots are real, we need the discriminant to be non-negative, so we require \((-4a + 12) \geq 0\). This ensures that every input falls into the category of either distinct real roots or a repeated real root. The first step in solving any quadratic problem is usually to calculate and analyze this discriminant. This directs us towards the potential conditions under which the equation can have real solutions.

Understanding Real Roots in Context

Real roots are simply the points where the graph of the quadratic equation intersects the x-axis. They are essential in many practical applications like physics, where they can represent points in time or space. In the context of equations, when solving for real roots, the focus is on checking whether the calculated values actually exist on the number line.In our given problem, the condition was not just about having real roots. They also needed to be less than 3. By setting up the inequality and considering a test value, such as \(x = 3\), we thoroughly check if the roots reside in the desired interval. This involves substituting into the equation and analyzing the inequality \(a^2 - 5a + 6 < 0\) to see which values of \(a\) confine the roots to be less than 3. Real roots can often lead to tangible outcomes, such as maximum profits in business, distances or depths in engineering, and much more.

Solving Quadratic Inequalities

Quadratic inequalities often arise when you need to determine intervals on the real number line. They can be solved similarly to equations but offer an extra layer of challenge. We aim to find where the inequality holds true.The steps usually start by solving the quadratic as an equation, \((a - 2)(a - 3) = 0\), which gives the critical points \(a = 2\) and \(a = 3\). We then determine the intervals: \((-\infty, 2)\), \((2, 3)\), and \((3, \infty)\), and test sample values from each interval to check where the inequality \((a - 2)(a - 3) < 0\) holds.This is done using sign analysis or a table, noting that the sign of the product of factors changes at critical points. For our problem, the inequality holds between the critical points \((2, 3)\). Thus, our final solution combines this with the condition derived from the discriminant to find \(2 < a \leq 3\). This forms the complete set of solutions that meet all required conditions.