Problem 50 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation \(\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1\) has (A) no solution (B) one solution (C) two solutions (D) more than two solutions

Step-by-Step Solution

Answer

The equation has two solutions.

Step 1: Set a substitution

Let us introduce a substitution to simplify the equation. Let \( \sqrt{x-1} = t \). This implies that \( x = t^2 + 1 \). The domain condition here is \( t \geq 0 \), as the square root cannot be negative.

Step 2: Rewrite the equation

Substitute \( x = t^2 + 1 \) into the original equation: \[ \sqrt{(t^2 + 1)+3 - 4t} + \sqrt{(t^2 + 1)+8 - 6t} = 1 \]Simplify each term:- First term: \( \sqrt{t^2 - 4t + 4} = \sqrt{(t-2)^2} \)- Second term: \( \sqrt{t^2 - 6t + 9} = \sqrt{(t-3)^2} \)

Step 3: Evaluate the square roots

The expressions under the square root, \( \sqrt{(t-2)^2} \) and \( \sqrt{(t-3)^2} \), as squares, simplify to absolute values:- \( \sqrt{(t-2)^2} = |t-2| \)- \( \sqrt{(t-3)^2} = |t-3| \)

Step 4: Solve the absolute values equation

Substitute the simplified terms into the equation:\[ |t-2| + |t-3| = 1 \]This equation implies that the sum of the distances from \( t \) to 2 and 3 equals 1.

Step 5: Analyze the cases from absolute values

Consider the possible values of \( t \): - For \( t < 2 \): \(|t-2| = 2-t\) and \(|t-3| = 3-t\). The equation \( (2-t) + (3-t) = 1 \) gives no solution, as it simplifies to \( 5 - 2t = 1 \), thus \( 2t = 4 \) and \( t = 2 \), which contradicts \( t < 2 \).- For \( 2 \leq t < 3 \): \(|t-2| = t-2\) and \(|t-3| = 3-t\). The equation \( (t-2) + (3-t) = 1 \) holds true, as it reduces to \( 1 = 1 \), so for this case, \( t \) is any value in \([2, 3)\).- For \( t \geq 3 \): \(|t-2| = t-2\) and \(|t-3| = t-3\). The equation \( (t-2) + (t-3) = 1 \) simplifies to \( 2t - 5 = 1 \), implying \( 2t = 6 \) and \( t = 3 \), which satisfies \( t \geq 3 \).

Step 6: Solution validity and conclusion

The solution \( t \in [2, 3] \) implies solutions for \( x \) as follows:- \( t = 2 \) gives \( x = 2^2 + 1 = 5 \).- \( t = 3 \) gives \( x = 3^2 + 1 = 10 \).Hence, each value of \( x \) satisfies the transformed equation, confirming that there are two solutions.

Key Concepts

Algebraic Equations

Algebraic equations are mathematical statements equating two algebraic expressions. They can involve variables, constants, and arithmetic operations. Solving algebraic equations often requires finding the value of the variable that makes the equation true. In the exercise, we start with a complex algebraic equation involving radical expressions and aim to simplify it using substitutions.

Understanding algebraic equations can involve several steps, including:

• Identifying like terms and simplifying expressions
• Using techniques like factoring or using algebraic identities
• Checking if the solution satisfies the original equation

Decomposing the equation into simpler forms can make finding the solution more manageable. In this exercise, substitution plays a crucial role, reducing the complexity of radical expressions and allowing us to solve it more efficiently.

Substitution Method

The substitution method is a powerful technique to simplify complex algebraic equations. By substituting one expression for a variable, we can transform the equation into a simpler form that is easier to manage. In the given exercise, we substitute \( \sqrt{x-1} = t \).

This change of variable is particularly useful when dealing with square roots or expressions that naturally yield to simplification. The steps typically involve:

• Determining a suitable expression related to the variable
• Replacing the original variable and rewriting the equation
• Simplifying the transformed equation and solving it

In our exercise, substituting allowed the equation involving radicals to be expressed as simpler quadratic forms, which we then solved using basic algebraic principles.

Absolute Value Equations

Absolute value equations express the non-negative magnitude of a number without considering its sign. Mathematically, the absolute value of a number \( x \) is denoted as \( |x| \).

In this exercise, after substitution and simplification, we deal with the absolute value equation \( |t-2| + |t-3| = 1 \). Solving such equations typically involves considering various cases based on the critical points that nullify absolute values:

• Examining different intervals, e.g., \( t < 2 \), \( 2 \leq t < 3 \), and \( t \geq 3 \)
• Formulating equations for each interval without absolute value signs
• Determining which intervals yield valid solutions and ensuring they comply with the original problem

The absolute value equation in this exercise highlights how distances between values \( t \) and defined points on a number line must add to a constant value, in this case, 1, leading us to explore separate potential scenarios for solutions.

Domain Restriction

Domain restriction is crucial when managing functions involving square roots, as it defines the permissible set of input values. In mathematical terminology, the domain is the set of all possible input values that make the function defined. Practically, it involves ensuring that values under a square root are non-negative.

In the exercise, the initial step involves using the substitution \( \sqrt{x-1} = t \), leading to a domain condition \( t \geq 0 \). This ensures that the underlying expression \( x-1 \geq 0 \) is always satisfied:

• Simplifying to find \( x \geq 1 \)
• Makes sure that any substituted or simplified equations adhere to this restriction
• Ensures solutions to the equation are realistic and valid

Understanding domain restrictions helps avoid invalid calculations and focuses on realistic solution sets for algebraic equations that involve radicals.