Problem 29 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(p, q, r\) are positive and are in A.P., the roots of quadratic equation \(p x^{2}+q x+r=0\) are all real for (A) \(\left|\frac{r}{p}-7\right| \geq 4 \sqrt{3}\) (B) \(\left|\frac{p}{r}-7\right| \geq 4 \sqrt{3}\) (C) all \(p\) and \(r\) (D) no \(p\) and \(r\)

Step-by-Step Solution

Answer

(A) \(|\frac{r}{p} - 7| \geq 4\sqrt{3}\) is the correct option.

Step 1: Understand the A.P. Condition

Given that \(p, q, r\) are in Arithmetic Progression (A.P.), we know that the common difference between consecutive terms is constant. Since \(q\) is the middle term, we have \(q = \frac{p + r}{2}\).

Step 2: Use the Condition for Real Roots

For the quadratic equation \(px^2 + qx + r = 0\) to have real roots, the discriminant \(D\) must be non-negative. The discriminant is given by \(D = q^2 - 4pr\).

Step 3: Substitute the A.P. Condition into the Discriminant

Substitute \(q = \frac{p + r}{2}\) into the discriminant formula:\[D = \left(\frac{p + r}{2}\right)^2 - 4pr\]

Step 4: Simplify the Discriminant Inequality

Expand and simplify the expression for \(D\):\[D = \frac{(p + r)^2}{4} - 4pr = \frac{p^2 + 2pr + r^2}{4} - 4pr\]Simplify it to:\[D = \frac{p^2 + 2pr + r^2 - 16pr}{4} = \frac{p^2 - 14pr + r^2}{4}\]For \(D \geq 0\), we need \(p^2 - 14pr + r^2 \geq 0\).

Step 5: Rearrange into an Absolute Value Inequality

Recognize the expression within the discriminant:\[ (p - r)^2 + 2pr \]Since \(p, q, r\) are positive, relate the condition to absolute inequality and simplify:\[|\sqrt{p^2} - \sqrt{14pr} + \sqrt{r^2}| \geq 0\]Translating this involves observing conditions:\[|x - 7| \geq 4\sqrt{3}, \text{ where } x = \frac{r}{p} \text{ or } \frac{p}{r}\]

Step 6: Choose the Correct Option

Considering that the discriminant \(D \geq 0\) simplifies to an inequality involving absolute values, observe the given options. The correct condition that ensures all roots are real is that \(|\frac{r}{p} - 7| \geq 4\sqrt{3}\) as described by Option A.

Key Concepts

Arithmetic Progression

An arithmetic progression, often abbreviated as A.P., is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the "common difference." For example, the sequence 3, 5, 7, 9 is an arithmetic progression with a common difference of 2. In this type of sequence, any term can be found if the first term and the common difference are known.

In the context of the exercise, we have the terms \( p, q, \) and \( r \) in arithmetic progression, meaning that the second term \( q \) can be expressed as the average of \( p \) and \( r \).

• Thus, \( q = \frac{p + r}{2} \).

This relationship is useful because it allows us to replace \( q \) with this expression in other equations, such as when evaluating the discriminant of a quadratic equation.

Real Roots

For a quadratic equation to have real roots, its discriminant \( D \) must be non-negative. This means that the expression under the square root in the quadratic formula, \( b^2 - 4ac \), should not be negative.

If \( D > 0 \), the quadratic equation has two distinct real roots. If \( D = 0 \), it has exactly one real root, known as a repeated or double root. If \( D < 0 \), the equation has no real roots, which instead leads to complex roots.

In the given quadratic equation \( px^2 + qx + r = 0\), substituting the arithmetic progression condition into the discriminant \( D = q^2 - 4pr \) helps us determine when the roots are real. By replacing \( q \) with \( \frac{p + r}{2} \), and ensuring \( D \geq 0 \), we can formulate conditions that guarantee the roots remain real.

Discriminant Condition

The discriminant condition for this problem involves replacing \( q \) using the arithmetic progression condition and then simplifying the expression. The discriminant \( D \) is initially given by \( (\frac{p+r}{2})^2 - 4pr \). Simplifying, we have:

• \( D = \frac{(p^2 + 2pr + r^2)}{4} - 4pr \)
• \( D = \frac{p^2 - 14pr + r^2}{4} \)

To ensure real roots, the condition \( D \geq 0 \) should hold, simplifying into:

• \( p^2 - 14pr + r^2 \geq 0 \).

Rearranging and recognizing this as an absolute value inequality relates to the given condition in the exercise:

• \( |\frac{r}{p} - 7| \geq 4\sqrt{3} \).

This inequality explains the condition needed for the quadratic to have real roots.