If $x - \frac{1}{x} = 4$, then what is the value of $x^2 + \fr... | Mathematics MCQ | Examoogle

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If $x - \frac{1}{x} = 4$ , then what is the value of $x^2 + \frac{1}{x^2}$ ?

Correct Answer

📐 MA • Math Direct FormulaMathematicsAlgebra

Option C

Quick Summary:

For the equation $x - 1/x = n$ , the value of $x^2 + 1/x^2$ is simply $n^2 + 2$ . Here, $4^2 + 2 = 16 + 2 = 18$ .

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📐MAMath SolutionDirect Formula

📋 Given

The expression given is x - (1/x) = 4.

🔢 Formula Used

(a−b)2=a2+b2−2ab(a - b)^2 = a^2 + b^2 - 2ab(a−b)2=a2+b2−2ab

⚡ Exam Hall Shortcut / Speed Trick

For the equation $x - 1/x = n$ , the value of $x^2 + 1/x^2$ is simply $n^2 + 2$ . Here, $4^2 + 2 = 16 + 2 = 18$ .

⚠️ Common Student Trap / Pitfall

Many students mistakenly use $n^2 - 2$ instead of $n^2 + 2$ , which is only applicable when given $x + 1/x = n$ .

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Square both sides of the given equation

To introduce the $x^2$ terms, square both sides of the equation $x - \frac{1}{x} = 4$ .

(x−1x)2=42(x - \frac{1}{x})^2 = 4^2(x−x1​)2=42

Expand using algebraic identity

Apply the identity $(a - b)^2 = a^2 - 2ab + b^2$ where $a = x$ and $b = \frac{1}{x}$ .

x2−2(x)(1x)+1x2=16x^2 - 2(x)(\frac{1}{x}) + \frac{1}{x^2} = 16x2−2(x)(x1​)+x21​=16

Simplify and solve

The middle term $2(x)(\frac{1}{x})$ simplifies to $2$ . Adding $2$ to both sides gives the final value.

x2−2+1x2=16  ⟹  x2+1x2=18x^2 - 2 + \frac{1}{x^2} = 16 \implies x^2 + \frac{1}{x^2} = 18x2−2+x21​=16⟹x2+x21​=18

✅

C is correct because squaring the given expression leads to $x^2 - 2 + 1/x^2 = 16$ , which results in $x^2 + 1/x^2 = 18$ .

Core Concepts Used

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Algebraic Identities Square of binomials Reciprocal relations

💡 EXAM TIP

This identity is fundamental for solving complex polynomial expressions in competitive exams; remember the 'sign-flip' logic: if you are given a minus, you add 2 to the square; if you are given a plus, you subtract 2.

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