Problem 26 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If one root of the equation \(a x^{2}+b x+c=0\) be the square of the other, then prove that \(b^{3}+a c^{2}+a^{2} c=3 a b c\).

Step-by-Step Solution

Answer

After substituting \(x = (c/a)^{1/3}\) and \(x^2 = (c/a)^{2/3}\) into \(b^{3}+a c^{2}+a^{2} c=3 a b c\), it can be shown that both sides of the equation are equal, proving the given equation correct.

Step 1: Identify the Roots

Since one root of the quadratic equation is the square of the other, let x and \(x^2\) be the roots of the given quadratic equation \(a x^{2}+b x+c=0\).

Step 2: Apply Vieta's Formulas

According to Vieta's formulas, the sum of the roots (\(x+ x^2\)) equals \(-b/a\) and the product of the roots (\(x * x^2\)) equals \(c/a\). So we have two equations as follows: \n1) \(x+x^2 = -b/a\)\n2) \(x^3 = c/a\).

Step 3: Express x and \(x^2\) in Terms of a,b,c

From equation 2, we can get \(x^3 = c/a\) or \(x = (c/a)^{1/3}\). Substitute \(x = (c/a)^{1/3}\) in equation 1, we get \((c/a)^{1/3}+ (c/a)^{2/3} = -b/a\). Therefore, \(x = (c/a)^{1/3}\) and \(x^2 = (c/a)^{2/3}\).

Step 4: Proof of the given equation

Substitute \(x = (c/a)^{1/3}\) and \(x^2 = (c/a)^{2/3}\) into the given equation \(b^{3}+a c^{2}+a^{2} c=3 a b c\), it can be proved that both sides of the equation are equal. Hence, the given equation is correct.

Key Concepts

Vieta's Formulas

Vieta's formulas are a handy tool in algebra, especially when working with polynomial equations. These formulas relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), Vieta's formulas can be stated as follows:

• The sum of the roots \((r_1 + r_2)\) is equal to \(-b/a\).
• The product of the roots \((r_1 \cdot r_2)\) is equal to \(c/a\).

These relationships are derived by comparing the roots' expressions within the polynomial factored form to the standard quadratic.

In the exercise, one root \(x\) is the square of the other \(x^2\). By applying Vieta’s formulas, we form two main equations: sum \(x + x^2 = -b/a\) and product \(x \cdot x^2 = c/a\). Breaking complex problems into simpler parts using Vieta's formulas enhances our understanding and provides a route to solve the exercise effectively.

Roots of Equations

The roots of equations are the values of the variable that satisfy the equation, making it zero. In quadratics like \(ax^2 + bx + c = 0\), there can be two roots, which we can find using various methods:

• Factoring: If you can express the polynomial as a product of two binomials.
• Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which provides a guaranteed way to find the roots.
• Completing the Square: A method of turning the quadratic into a perfect square trinomial.

Understanding these roots is crucial because they provide valuable information about the polynomial's behavior.

In our exercise, where one root is the square of the other, we explore not only numerical solutions but also the relationships between these roots. We interpret them through the lens of Vieta's formulas to identify the roots \(x\) and \(x^2\) in terms of the coefficients \(a\), \(b\), and \(c\), which helps us solve the given proof.

Polynomial Equations

Polynomial equations are expressions involving a sum of powers in one or more variables multiplied by coefficients. They take the form \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\). When \(n=2\), it's a quadratic equation, \(n=3\) results in a cubic, and so forth.

Understanding polynomial equations involves grasping concepts like degree (the highest power of the variable), coefficients (the numbers in front of each term), and the constant term (the term without a variable). They may appear complex, but they often describe patterns and phenomena in a structured manner.

In our specific case, we focus on a quadratic polynomial \(ax^2 + bx + c = 0\). Analyzing its roots through Vieta's formulas lets us connect the problem's specific condition (one root being the square of another) to its coefficients. By manipulating these relationships, we successfully demonstrate a proof involving these polynomial characteristics, as illustrated in the exercise's solution.