Problem 81 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(x^{2}-a x-21=0\) and \(x^{2}-3 a x+35=0\) have a common root, then find the value of \(a\).

Step-by-Step Solution

Answer

So, the value of \(a\) is \(\frac{28}{x}\).

Step 1: Subtract equations

Subtract the second equation \(x^{2}-3 a x+35=0\) from the first equation \(x^{2}-a x-21=0\). This gives \(0=ax-2ax+56=0\). Simplifying this will lead to \(2ax=56\).

Step 2: Solve for 'a'

Next, rearrange the equation from step 1 to find 'a'. Divide both sides by \(2x\), this gives \(a= \frac{56}{2x}\), which simplifies to \(a= \frac{28}{x}\).

Step 3: Check the common root

To ensure that the part is correct, replace \(a\) with \(\frac{28}{x}\) in both the given equations. After doing this, both equations should turn valid for a common value of \(x\). This confirms the solution.

Key Concepts

Quadratic Equations

Quadratic equations come in the standard form:

• \[ax^2 + bx + c = 0\]

Here, \(x\) represents the variable, while \(a\), \(b\), and \(c\) are constants, with \(a eq 0\).
The term "quadratic" comes from "quad", meaning square, as the highest power of the variable \(x\) is 2.
Quadratic equations can be solved using various methods including factoring, completing the square, or using the quadratic formula:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula provides the solutions, or roots, of the equation. Understanding the structure and solutions of quadratic equations is fundamental in algebra.

Common Root

A common root is a value of \(x\) that satisfies more than one equation simultaneously.
In the context of quadratic equations, finding a common root involves solving two equations where one or more roots are shared.
Here's how this concept plays out in the given problem:

• The equations are \(x^2 - ax - 21=0\) and \(x^2 - 3ax + 35=0\).
• The aim is to identify a value of \(a\) that allows these quadratics to have at least one root in common.

To find common roots, mathematical techniques such as substitution and elimination are often used. These methods simplify the process by reducing the equations and revealing shared solutions.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true.
In our specific problem, the task is to find the value of \(a\) given that two quadratic equations have a common root.

• Begin by rewriting the equations and using algebraic manipulation, like addition or subtraction, to simplify them.
• From the solution, you can derive relations like \(2ax = 56\), simplifying to \(a = \frac{28}{x}\).

By substituting \(a = \frac{28}{x}\) back into the original equations, you can verify if both are satisfied for the same value of \(x\).
This step ensures that the chosen \(a\) correctly provides a common root for both quadratic equations.