Problem 128 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The values of ' \(a\) ' so that 6 lies between the roots of the equation \(x^{2}+2(a-3) x+9=0\), are (A) \(a>-\frac{3}{4}\) (B) \(a<-\frac{3}{4}\) (C) \(a>\frac{3}{4}\) (D) \(a<\frac{3}{4}\)

Step-by-Step Solution

Answer

The value of \( a \) that ensures 6 lies between the roots is \( a > \frac{21}{4} \). None of the given options directly match this.

Step 1: Understanding the problem

We need to determine the conditions on the parameter \( a \) such that the number 6 lies between the roots of the quadratic equation \( x^2 + 2(a-3)x + 9 = 0 \). This means if the roots are \( r_1 \) and \( r_2 \), then the statement \( r_1 < 6 < r_2 \) or \( r_2 < 6 < r_1 \) should hold.

Step 2: Identify the roots using Vieta’s formulas

For any quadratic equation \( ax^2 + bx + c = 0 \), Vieta's formulas tell us the sum and the product of the roots: \( r_1 + r_2 = -\frac{b}{a} \) and \( r_1 \times r_2 = \frac{c}{a} \). For our equation \( x^2 + 2(a-3)x + 9 = 0 \), we have \( b = 2(a-3) \) and \( c = 9 \). Therefore, \( r_1 + r_2 = -2(a-3) = -2a + 6 \) and \( r_1 \times r_2 = 9 \).

Step 3: Determine condition for 6 lying between the roots

For \( 6 \) to lie between the roots, the mean value \( \frac{r_1 + r_2}{2} \) must be equal to 6: \( \frac{-2a + 6}{2} = 6 \). Solving this gives us \( -2a + 6 = 12 \), leading to \( a = -3 \). Thus, \( 6 \) is the midpoint of the distance between the roots.

Step 4: Determining inequalities from quadratic property

For \( 6 \) to lie strictly between the roots, neither root should be exactly 6. This means the roots should satisfy \( f(6) > 0 \) where \( f(x) = x^2 + 2(a-3)x + 9 \). Substitute \( x = 6 \) to find \( f(6) = 36 + 12(a-3) + 9 \). This simplifies to \( f(6) = 12a - 63 \), and for 6 to lie between the roots, \( 12a - 63 > 0 \) or \( 12a > 63 \). Solving gives \( a > \frac{21}{4} \). Hence, \( a > \frac{21}{4} \) ensures 6 is strictly between the roots.

Key Concepts

Vieta's Formulas

Vieta's formulas are a set of equations that relate the coefficients of a polynomial to sums and products of its roots. They are particularly useful when solving quadratic equations. If we consider a standard quadratic equation of the form \( ax^2 + bx + c = 0 \), Vieta's formulas give us two main relationships for the roots \( r_1 \) and \( r_2 \):

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

These formulas allow us to find significant properties of the roots without actually solving the quadratic equation.

In the context of the problem, the equation is \( x^2 + 2(a-3)x + 9 = 0 \). Here, the coefficients \( b \) and \( c \) are \( 2(a-3) \) and \( 9 \), respectively. Applying Vieta's formulas:

• Sum of roots, \( r_1 + r_2 = -2(a-3) = -2a + 6 \)
• Product of roots, \( r_1 \times r_2 = 9 \)

This understanding helps when determining conditions for specific root properties, such as having a specified value situated between them.

Roots of Quadratic Equation

Roots of a quadratic equation are the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). These can be obtained either by factoring, using the quadratic formula, or employing methods like Vieta's formulas.

In the quadratic equation provided, \( x^2 + 2(a-3)x + 9 = 0 \), finding the roots involves understanding their behavior through the formulas we have. Since we don't need exact values for each root, Vieta's formulas suffice here.

The key task is to ensure that 6 lies between the roots \( r_1 \) and \( r_2 \).

To achieve this, recognize that:

• The average of the roots \( \frac{r_1 + r_2}{2} \) must equal to 6. Solving this gives \( -2a + 6 = 12 \), simplifying to \( a = -3 \). However, it does not fulfill the inequality condition.
• Instead, when combining the product property \( r_1 \times r_2 = 9 \), and requiring that the quadratic evaluated at x=6 is positive, you find \( a > \frac{21}{4} \). This ensures root spacing sufficient for 6 to lie strictly between them.

Understanding these ensures a deeper grasp of how the properties of quadratic structures impact their graph and root positions.

Inequalities

Inequalities involve mathematical expressions that compare different values. When working with quadratic equations, inequalities can help determine the nature and spacing of the roots.

In this exercise, recognizing that 6 should lie between the roots of the equation means solving not just for equalities, but also handling inequalities effectively. For the equation \( x^2 + 2(a-3)x + 9 = 0 \):

• First, let's remember that if 6 lies strictly between the roots, it should not equal either root.
• Hence, we solve for the condition \( f(6) > 0 \) in \( x^2 + 2(a-3)x + 9 \). This gives us \( 36 + 12(a-3) + 9 = 12a - 63 \).
• To satisfy the inequality \( 12a - 63 > 0 \), we solve \( 12a > 63 \) leading to \( a > \frac{21}{4} \).

This inequality ensures that the number 6 is strictly between the roots, as necessary for this quadratic setup. Understanding how inequalities interact with roots helps in determining more nuanced conditions for root placement relative to specific values.