Problem 108 - A Complete Resource Book in Mathematics for JEE Main

Question

If the equation \(x^{2}+\left[a^{2}-5 a+b+4\right] x+b=0\) has roots \(-5\) and 1 , where \([a]\) denotes the greatest integer less than or equal to \(a\), then the set of values of \(a\) is (A) \(\left(\frac{5-3 \sqrt{5}}{2}, \frac{5+3 \sqrt{5}}{2}\right)\) (B) \(\left(0, \frac{5+3 \sqrt{5}}{2}\right)\) (C) \(\left(-1, \frac{5-3 \sqrt{5}}{2}\right] \cup\left[\frac{5+3 \sqrt{5}}{2}, 6\right)\) (D) None of these

Step-by-Step Solution

Answer

The correct set of values is option (A): \( \left( \frac{5 - 3\sqrt{5}}{2}, \frac{5 + 3\sqrt{5}}{2} \right) \).

Step 1: Apply Vieta's Formulas

According to Vieta's formulas, the sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \(-\frac{b}{a}\), and the product is \(\frac{c}{a}\). In our case, the equation is \(x^2 + [a^2 - 5a + b + 4]x + b = 0\). Given roots are \(-5\) and \(1\).Sum of roots: \(-5 + 1 = -4 = -[a^2 - 5a + b + 4].\)Product of roots: \(-5 \times 1 = -5 = b.\)

Step 2: Substitute the Product of Roots

From the product of the roots, \(b = -5\). Substitute \(b\) into the sum of roots equation:\[-4 = -[a^2 - 5a + (-5) + 4] = -[a^2 - 5a - 1].\]

Step 3: Solve the Sum of Roots Equation

Rewriting the equation from the previous step, we have:\[a^2 - 5a - 1 = 4.\]Simplifying, we get:\[a^2 - 5a - 5 = 0.\]

Step 4: Solve the Quadratic Equation

We solve \(a^2 - 5a - 5 = 0\) using the quadratic formula, \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1, b = -5, c = -5\). Thus, \[a = \frac{5 \pm \sqrt{25 + 20}}{2} = \frac{5 \pm \sqrt{45}}{2}.\]

Step 5: Simplify the Roots

Simplify the expression \(\sqrt{45}\): \(\sqrt{45} = 3 \sqrt{5}\).Hence, the roots for \(a\) become \(\frac{5 \pm 3\sqrt{5}}{2}\). This implies the interval for \(a\) is between these two roots.

Step 6: Interpret the Integer Part

Recall that \([a]\) is the greatest integer less than or equal to \(a\). Therefore, for the equation to hold true, \(a\) must take values from the interval but cannot include integers as \([a]\) must affect the quartile inclusion properly.The correct interpretation of this interval is \(a \in \left( \frac{5 - 3\sqrt{5}}{2}, \frac{5 + 3\sqrt{5}}{2} \right)\).

Key Concepts

Vieta's Formulas

Vieta's formulas are a helpful set of equations used to relate the coefficients of a polynomial to sums and products of its roots. These formulas are particularly useful when working with quadratic equations, which are of the form \( ax^2 + bx + c = 0 \). For quadratic equations:

• The sum of the roots \( (x_1 + x_2) \) is given by \(-\frac{b}{a}\).
• The product of the roots \( (x_1 \cdot x_2) \) is given by \(\frac{c}{a}\).

In the given problem, we have a quadratic equation \[ x^2 + (a^2 - 5a + b + 4)x + b = 0 \]and we're told that the roots are \(-5\) and \(1\). By applying Vieta's formulas:

• Sum of the roots: \(-5 + 1 = -4 = -[a^2 - 5a + b + 4].\)
• Product of the roots: \(-5 \times 1 = -5 = b.\)

These equations help us draw a direct link between the roots and the coefficients of the polynomial equation, allowing us to derive the set of possible values of \(a\).

Greatest Integer Function

The Greatest Integer Function, often denoted as \([a]\), is a mathematical function that rounds down a real number to the nearest integer less than or equal to that number. It's also known as the "floor" function. Here's how it operates:

• For any real number \(a\), \([a]\) is the greatest integer less than or equal to \(a\).
• For example, \([3.7] = 3\), \([-2.3] = -3\), and \([5] = 5\).

In the context of the given equation, \([a]\) influences the coefficient of \(x\). Thus, when finding the range for \(a\), it's crucial that \([a]\) appropriately adjusts the interval, ensuring it excludes integers within the interval \(\left( \frac{5 - 3\sqrt{5}}{2}, \frac{5 + 3\sqrt{5}}{2} \right)\).This concept is key to solving the problem accurately because it adjusts the values \(a\) can take such that the equation remains valid.

Roots of Polynomial Equations

Understanding the roots of polynomial equations, especially quadratics, is foundational in algebra. When solving for roots, or solutions of the equation, we find the values that make the equation true. For a quadratic equation \(ax^2 + bx + c = 0\), the roots can be solved using:

• The quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• By factoring, when possible, which involves expressing the equation as \((x - p)(x - q) = 0\), implying roots \(p\) and \(q\).

In this solution, we use the quadratic formula to solve the equation \[a^2 - 5a - 5 = 0\].Using the formula, the roots for \(a\) become \[\frac{5 \pm 3\sqrt{5}}{2}\],which provide the interval where \(a\) can be found between these two values. Simplifying the roots gives us a clear interval, allowing us to focus on specific values of \(a\) that satisfy all given conditions of the problem.