Problem 54 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(x^{2}-(a+b+c) x+(a b+b c+c a)=0\) has imaginary roots, where \(a, b, c \in R^{+}\), then \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) (A) can be the sides of a triangle (B) cannot be the sides of a triangle (C) nothing can be said (D) None of these

Step-by-Step Solution

Answer

(A) can be the sides of a triangle

Step 1: Understanding Imaginary Roots

For the quadratic equation \(x^2-(a+b+c)x+(ab+bc+ca)=0\) to have imaginary roots, the discriminant must be negative. The discriminant \(\Delta\) for the equation \(ax^2+bx+c=0\) is given by \(b^2-4ac\). In our case, \(a=1, b=-(a+b+c), c=ab+bc+ca\). According to the question, \(\Delta < 0\).

Step 2: Calculating the Discriminant

The discriminant for the given quadratic equation is \((a+b+c)^2 - 4(ab+bc+ca)\). This simplifies to \(a^2+b^2+c^2 - 2(ab+bc+ca)\). The condition for imaginary roots is \(a^2+b^2+c^2 < 2(ab+bc+ca)\).

Step 3: Analyzing the Condition

The inequality \(a^2+b^2+c^2 < 2(ab+bc+ca)\) is equivalent to \((a-b)^2 + (b-c)^2 + (c-a)^2 < 0\). Since a square of a real number is always non-negative, this implies that if the inequality holds, then it should be true that each term \((a-b)^2, (b-c)^2, (c-a)^2\) is zero, meaning \(a=b=c\).

Step 4: Determining Validity for Triangle Sides

If \(a=b=c\), then \(\sqrt{a}=\sqrt{b}=\sqrt{c}\). These can form the sides of an equilateral triangle because the triangle inequality \(\sqrt{a} + \sqrt{b} > \sqrt{c}\) holds due to equal addition of any two sides being greater than the third. Therefore, \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) can indeed be the sides of a triangle.

Key Concepts

Triangle Inequality

The triangle inequality is a fundamental principle that helps determine whether three lengths can form the sides of a triangle. It states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This ensures that the sides are able to meet and create a closed shape.

When examining potential triangle sides, remember this rule:

• If you have three potential sides, like \(a, b,\) and \(c\), then:
• \(a + b > c\)
• \(a + c > b\)
• \(b + c > a\)

In our exercise, we consider if \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) can be these sides. If each of these conditions is met, the lengths form a valid triangle. Specifically, the side lengths being equal satisfies this inequality naturally, as in the case of an equilateral triangle where each side is equal in length.

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree 2. It follows the standard form \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constants, with \(a eq 0\). The solutions, or roots, of a quadratic equation can offer significant insights into the properties of the function it represents.

To understand it step-by-step:

• "Quadratic" comes from "quad" meaning square, as the highest power is 2.
• The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of \(a\).

Roots can be real or imaginary based on the discriminant of the equation. In our example, the focus is on when the roots are imaginary, which provides a specific condition related to the equation's parameters.

Discriminant

The discriminant is a special part of the quadratic equation that helps determine the nature of its roots. It is found in the formula \(b^2 - 4ac\) for a quadratic equation \(ax^2 + bx + c = 0\). This value provides insight into the solutions without needing to solve the equation fully.

Here's what the discriminant tells us:

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If the discriminant is zero, the quadratic equation has exactly one real root (a repeated root).
• If the discriminant is negative, the quadratic equation has two complex (or imaginary) roots.

In the provided exercise, a negative discriminant indicates imaginary roots, crucial for analyzing if the given terms can represent triangle sides. Hence, the condition \((a-b)^2 + (b-c)^2 + (c-a)^2 < 0\) led us to concluding that \(a = b = c\), which satisfied the triangle inequality as well.