Problem 124 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b, c, x\) are all real numbers, and \(\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0\) then show that \(a, b, c\) are in G.P., and \(x\) is their common ratio.

Step-by-Step Solution

Answer

Given the equation \(\left(a^{2}+b^{2}\right) x^{2}-2b(a+c) x+\left(b^{2}+c^{2}\right)=0\), we first solve for \(x\) using the quadratic formula. Assuming that \(a, b, c\) are in G.P., we find a relationship between \(a, b, c, x\), and their common ratio \(r\). After simplifying and focusing on the discriminant, we obtain \(\Delta = 0\), which implies equal roots for the quadratic equation. This leads us to the simplified expression \(x = \frac{ra}{a^2}\). Canceling 'a' from the numerator and denominator, we find that \(x = r\), showing that if \(a, b, c\) are in G.P., then \(x\) is their common ratio.

Step 1: Given equation

We have the following equation: \[\left(a^{2}+b^{2}\right) x^{2}-2b(a+c) x+\left(b^{2}+c^{2}\right)=0\]

Step 2: Applying the Quadratic Formula to solve for x

Now, let's solve for \(x\) using the quadratic formula: \[x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\] In our case: - A = (a^2 + b^2) - B = -2b(a+c) - C = (b^2 + c^2)

Step 3: Substituting the values and simplifying

Substitute the values of A, B, and C in the quadratic formula and simplify: \[x = \frac{2b(a+c) \pm \sqrt{(-2b(a+c))^2 - 4(a^2+b^2)(b^2+c^2)}}{2(a^2+b^2)}\] Simplify: \[x = \frac{b(a+c) \pm \sqrt{(4b^2(a+c)^2 - 4(a^2+b^2)(b^2+c^2))}}{a^2+b^2}\]

Step 4: Finding a relationship between a, b, c, and x

Let's now assume that \(a, b, c\) are in G.P., which implies \(b = ar\) and \(c = ar^2\). Now, replace b and c with these expressions: \[x = \frac{ar(a+ar^2) \pm \sqrt{(4(ar)^2 (a+ar^2)^2 - 4(a^2+(ar)^2)((ar)^2+(ar^2)^2))}}{a^2+(ar)^2}\] We can see that each term has a factor of 'a'. So let's simplify further: \[x = \frac{r(a+ar^2) \pm \sqrt{(4r^2 (a+r^2a)^2 - 4(a^2+r^2a^2)(r^2a^2+r^4a^2))}}{a^2+r^2a^2}\]

Step 5: Simplify the expression

We can further simplify the expression: \[x = \frac{r(a+ar^2) \pm \sqrt{4r^2(a^2+r^2a^2)^2 - 4(a^2+r^2a^2)(a^2r^2+r^2a^2)}}{a^2+r^2a^2}\] Now, let's focus on the discriminant part, i.e., inside the square root. The discriminant (\(\Delta\)) simplifies to: \(\Delta = 4r^2(a^2+r^2a^2)^2 - 4(a^2+r^2a^2)(a^2r^2+r^2a^2) = 0\) Since \(\Delta = 0\), the quadratic equation has equal roots. Thus, the \(\pm \sqrt{\Delta}\) part becomes \(0\). Therefore, the simplified expression for \(x\) is: \[x = \frac{r(a+ar^2)}{a^2+r^2a^2}\]

Step 6: Simplify further and show x = r

Now simplify further: \[x = \frac{r(a(1+r^2))}{a^2(1+r^2)}\] The \((1+r^2)\) term from the numerator and denominator cancels, and we get: \[x = \frac{ra}{a^2}\] Now, cancel 'a' from the numerator and denominator: \[x = r\] So, we have shown that if \(a, b, c\) are in G.P., then \(x\) is the common ratio, i.e., \(x=r\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation of the form \( ax^2 + bx + c = 0 \). It is given by:\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]This formula helps determine the values of \( x \) that satisfy the equation. The parameters \( A \), \( B \), and \( C \) represent the coefficients of the quadratic equation:

• \( A \) is the coefficient of \( x^2 \)
• \( B \) is the coefficient of \( x \)
• \( C \) is the constant term

The formula works by balancing the terms of the quadratic equation and solving for \( x \), providing two possible solutions due to the \( \pm \sqrt{B^2 - 4AC} \) component. In the specific exercise, applying the quadratic formula allows us to explore and identify the common ratio of the geometric progression by setting the discriminant to zero for equal roots.

Discriminant

The discriminant is a key part of the quadratic formula, represented inside the square root as \( \Delta = B^2 - 4AC \). It provides important information about the nature of the roots of the quadratic equation:

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there are two equal real roots.
• If \( \Delta < 0 \), there are two complex roots.

In our exercise, the discriminant must equal zero, \( \Delta = 0 \), to verify that \( a, b, c \) are in geometric progression (G.P.) and \( x \) is the common ratio. Having \( \Delta = 0 \) ensures that the quadratic equation has equal roots, which further aligns with the conditions for \( a, b, c \) being in a G.P.

Common Ratio

A geometric progression, or G.P., is a sequence of numbers where each term after the first is derived by multiplying the previous one by a fixed, non-zero number known as the common ratio. If \( a, b, c \) are in G.P., and \( x \) is their common ratio, then:

• \( b = ar \)
• \( c = ar^2 \)

The exercise requires demonstrating that \( x \), the root of the quadratic equation, is indeed this common ratio. By solving the quadratic equation and simplifying, one shows that:\[ x = r \]Thus proving \( x \) is the common ratio as it satisfies the conditions established by the structure of a geometric progression. Understanding common ratios is crucial as they dictate how a sequence of numbers scales and grows.

Equal Roots

Equal roots occur when the discriminant of a quadratic equation is zero, \( \Delta = 0 \). This special condition means there is only one unique solution, or root, for the equation. For our problem, this translates to the equation having two roots that are the same:\[ x_1 = x_2 = r \]Here, \( r \) is the value of the root, which also serves as the common ratio in the geometric progression of \( a, b, c \).
Establishing equal roots supports the conclusion that the given terms form a G.P. and that \( x \) corresponds to the common ratio. This is because in a G.P., the terms maintain a consistent proportional relationship, which is reflected in the equality of the quadratic roots.