Problem 20 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a x^{2}+b x+c=0, a \neq 0, a, b, c \in R\) has distinct real roots in \((1,2)\) then \(a\) and \(5 a+2 b+c\) have (A) same sign (B) opposite sign (C) not determined (D) None of these

Step-by-Step Solution

Answer

The signs of \(a\) and \(5a + 2b + c\) are opposite (Option B).

Step 1: Understand the Problem

We are given a quadratic equation \(ax^2 + bx + c = 0\) with distinct real roots in the interval (1,2). We need to determine the relationship of signs between \(a\) and \(5a + 2b + c\).

Step 2: Conditions for Distinct Real Roots

A quadratic equation has distinct real roots if its discriminant \(b^2 - 4ac > 0\). Additionally, the roots must both lie within the interval (1,2).

Step 3: Use the Quadratic Roots in Interval Condition

For roots in (1,2), the value of the quadratic at 1 and 2 should satisfy \(a(1)^2 + b(1) + c > 0\) and \(a(2)^2 + b(2) + c < 0\). This gives us \(a + b + c > 0\) and \(4a + 2b + c < 0\).

Step 4: Analyze the Inequalities

From the inequalities \(a + b + c > 0\) and \(4a + 2b + c < 0\), we find constraints on \(a, b,\) and \(c\). Subtract the first inequality from the second to isolate terms: \((4a + 2b + c) - (a + b + c) < 0\), simplifying to \(3a + b < 0\).

Step 5: Relationship Between a and 5a + 2b + c

Now, calculate \(5a + 2b + c\). We know from step 3 that \(4a + 2b + c < 0\), thus since \(4a + 2b < -c\), adding \(a\) leads to \(5a + 2b + c < a < 0\) if \(a > 0\). Therefore, \(5a + 2b + c\) would be negative, indicating opposite signs if \(a > 0\). The same argument holds considering \(a < 0\).

Step 6: Conclusion

From the analysis, if \(a > 0\) or \(a < 0\), \(5a + 2b + c\) is the opposite sign of \(a\). Hence, \(a\) and \(5a + 2b + c\) have opposite signs.

Key Concepts

Discriminant for real roots

To determine if a quadratic equation has real roots, we use the concept of the discriminant. The discriminant is a part of the quadratic formula, represented as \(b^2 - 4ac\). This value can tell us about the nature of the roots without actually finding them.

For a quadratic equation \(ax^2 + bx + c = 0\), the condition for real and distinct roots is when the discriminant is greater than zero, i.e., \(b^2 - 4ac > 0\). This indicates that both roots are real numbers and not equal.

• If \(b^2 - 4ac > 0\), the roots are real and distinct.
• If \(b^2 - 4ac = 0\), there is exactly one real root (also called a repeated root).
• If \(b^2 - 4ac < 0\), the roots are complex and imaginary.

Understanding this part of quadratic equations helps evaluate the potential solution space, particularly when determining where the roots might lie within a given interval.

Inequalities in quadratic roots

When dealing with quadratic equations, it's crucial to understand how the roots behave within certain intervals. In this context, the roots were specified to be within (1,2). To ensure that both roots of the quadratic equation indeed lie within this interval, two inequalities are set up by evaluating the quadratic expression at boundaries of the interval.

For the quadratic \(ax^2 + bx + c = 0\), if the roots fall in the interval, the equation must satisfy specific conditions at \(x=1\) and \(x=2\):

• At \(x=1\), the condition \(a + b + c > 0\) ensures the quadratic is positive just beyond the lower end of the interval.
• At \(x=2\), the condition \(4a + 2b + c < 0\) ensures the quadratic is negative just beyond the upper end of the interval.

By analyzing these inequalities, we determine the possible signs and meet the requirement of the roots existing within the specified region.

Sign analysis of coefficients

The relationship between the coefficients in a quadratic equation can reveal a lot about the equation's graph and its roots. Specifically, analyzing the sign of the leading coefficient \(a\) and its combinations with other coefficients can help determine the parabola's direction and roots placement.

An important inference in this problem involves the expression \(5a + 2b + c\). The task was to determine its sign relative to that of \(a\).

• The provided inequalities \(a + b + c > 0\) and \(4a + 2b + c < 0\) strongly influence the calculation.
• By manipulating these, we know \(3a + b < 0\). Expanding further, the expression \(5a + 2b + c\) also results opposite in sign to \(a\) because after adding \(a\) to \(4a + 2b + c < 0\), it remains negative.

This analysis methodically establishes that the terms have opposite signs, indicating the solution choice where \(a\) and \(5a + 2b + c\) always appear with opposite signs.