Problem 130 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The values of \(a\) for which each one of the roots of \(x^{2}-4 a x+2 a^{2}-3 a+5=0\) is greater than 2 , are (A) \(a \in(1, \infty)\) (B) \(a=1\) (C) \(a \in(-\infty, 1)\) (D) \(a \in(9 / 2, \infty)\)

Step-by-Step Solution

Answer

(A) \(a \in(1, \infty)\)

Step 1: Understand the Inequality Condition

To ensure both roots of the quadratic equation are greater than 2, we need to start with the conditions: each root, let's say \(\alpha\) and \(\beta\), should be greater than 2, i.e. \(\alpha > 2\) and \(\beta > 2\). We will use these inequalities to find the possible values of \(a\).

Step 2: Express Root Conditions Using Vieta's Formulas

For a quadratic equation \(ax^2 + bx + c = 0\), Vieta's formulas state that \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha \beta = \frac{c}{a}\). Here, the sum of the roots is \(4a\), and the product is \(2a^2 - 3a + 5\) (as the equation is normalized: \(x^2 - 4ax + (2a^2 - 3a + 5) \)). Thus, the root conditions become: \(\alpha + \beta > 4\) and \(\alpha \beta > 4\).

Step 3: Evaluate Root Sum Condition

From Vieta's formulas, substituting \(\alpha + \beta > 4\) yields the inequality: \(4a > 4\). This simplifies to \(a > 1\).

Step 4: Evaluate Root Product Condition

For the root product, we have the inequality \((2a^2 - 3a + 5) > 4\). Simplifying yields: \(2a^2 - 3a + 5 > 4\), or \(2a^2 - 3a + 1 > 0\). To solve this quadratic inequality, find the roots of the equation \(2a^2 - 3a + 1 = 0\) using the quadratic formula.

Step 5: Solve the Quadratic Inequality

The roots are found using the quadratic formula: \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), \(c = 1\). This results in the roots \(a = 1\) and \(a = \frac{1}{2}\). The quadratic \(2a^2 - 3a + 1\) is positive outside the interval \(\left(\frac{1}{2}, 1\right)\).

Step 6: Combine Both Conditions

We need \(a > 1\) from the sum of roots condition and \(a otin \left(\frac{1}{2}, 1\right)\) from the product root condition. Therefore, combining these, \(a > 1\).

Step 7: Choose the Correct Option

The valid set of \(a\) values where both roots are greater than 2 is \(a > 1\). Hence, the correct option is (A) \(a \in(1, \infty)\).

Key Concepts

Vieta's Formulas

Vieta's formulas provide a convenient way to connect the coefficients of a quadratic equation to its roots. For the quadratic equation \(ax^2 + bx + c = 0\), these formulas tell us:

• The sum of the roots \(\alpha + \beta = -\frac{b}{a}\)
• The product of the roots \(\alpha \beta = \frac{c}{a}\)

This means, without knowing the roots directly, you can ascertain their sum and product just by inspecting the coefficients of the polynomial. This technique becomes very handy when solving problems that involve relationships between the roots, such as finding inequalities the roots must satisfy.
In this specific example, the given quadratic equation is \(x^2 - 4ax + (2a^2 - 3a + 5)\). Applying Vieta's formulas here, we derive that the sum of the roots is \(4a\) and the product is \(2a^2 - 3a + 5\). To ensure the roots are greater than a specific value (in our case, 2), these derived expressions can be used to form inequalities. Furthermore, solving these inequalities helps in determining the acceptable set of values for \(a\).

Roots of Quadratic Equations

Identifying the roots of quadratic equations is crucial because they are the points where the quadratic expression equals zero. These roots can be found using various methods, such as factoring, completing the square, or using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation \(ax^2 + bx + c = 0\). The expression inside the square root, \(b^2 - 4ac\), is known as the discriminant and determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root, also known as a repeated or double root.
• If negative, the roots are complex and not real numbers.

In your exercise, roots were calculated using the quadratic formula, leading to \(a = 1\) and \(a = \frac{1}{2}\). These roots are used to solve quadratic inequalities, shaping the range of acceptable values for \(a\). Understanding these roots is key to deducing where our quadratic function is above or below a certain value.

Quadratic Inequalities

Quadratic inequalities involve expressions that contain a quadratic in opposition with a number or zero. The goal is to identify the set of values for which the inequality holds true, often producing a range instead of specific numbers.
To solve a quadratic inequality, follow these steps:

• Move all terms to one side of the inequality to compare the expression to zero.
• Find the roots using known methods like factoring or the quadratic formula.
• Use these roots to divide the number line into intervals.
• Test a value from each interval to see if it satisfies the inequality.
• Identify which intervals work and form your solution set.

In the exercise, the inequality \(2a^2 - 3a + 1 > 0\) was solved by determining that the function is positive outside the interval \(\left(\frac{1}{2}, 1\right)\). This process confirmed that for \(a > 1\), both roots of your quadratic equation are greater than 2, fitting the conditions specified in the problem statement. Solving quadratic inequalities like these helps in understanding the behavior of the quadratic function over various intervals.