If $ x + \frac{1}{x} = 5 $, then $ x^2 + \frac{1}{x^2} = ? $ | Mathematics MCQ | Examoogle

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If $x + \frac{1}{x} = 5$ , then $x^2 + \frac{1}{x^2} =?$

Correct Answer

📐 MA • Math Direct FormulaMathematicsPrevious Year Paper 2024

Option C

Quick Summary:

For $x + \frac{1}{x} = k$ , the value of $x^2 + \frac{1}{x^2}$ is always $k^2 - 2$ . Here, $5^2 - 2 = 25 - 2 = 23$ .

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📐MAMath SolutionDirect Formula

📋 Given

x $+ \frac{1}{x}$ = 5

🔢 Formula Used

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

⚡ Exam Hall Shortcut / Speed Trick

For $x + \frac{1}{x} = k$ , the value of $x^2 + \frac{1}{x^2}$ is always $k^2 - 2$ . Here, $5^2 - 2 = 25 - 2 = 23$ .

⚠️ Common Student Trap / Pitfall

Many students mistakenly square 5 and assume the answer is 25, forgetting to subtract the 2 arising from the expansion $(a+b)^2 = a^2 + b^2 + 2ab$ .

🔢 Step-by-Step Solution

Write down the given equation

We are given the value of the algebraic expression $x + \frac{1}{x} = 5$ .

x+1x=5x + \frac{1}{x} = 5x+x1​=5

Squaring both sides

Square both sides of the equation to introduce the quadratic terms $x^2$ and $\frac{1}{x^2}$ .

(x+1x)2=52(x + \frac{1}{x})^2 = 5^2(x+x1​)2=52

Expand the left hand side

Using the algebraic identity $(a+b)^2 = a^2 + b^2 + 2ab$ , expand the left side where $a = x$ and $b = \frac{1}{x}$ .

x2+1x2+2(x)(1x)=25x^2 + \frac{1}{x^2} + 2(x)(\frac{1}{x}) = 25x2+x21​+2(x)(x1​)=25

Simplify and calculate

Since $x \times \frac{1}{x} = 1$ , the equation simplifies to $x^2 + \frac{1}{x^2} + 2 = 25$ .

x2+1x2=25−2=23x^2 + \frac{1}{x^2} = 25 - 2 = 23x2+x21​=25−2=23

✅

C is correct because squaring $x + \frac{1}{x} = 5$ gives $x^2 + \frac{1}{x^2} + 2 = 25$ , which yields $23$ .

Core Concepts Used

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Algebraic Identities Squaring Binomials Reciprocal Equations

💡 EXAM TIP

This same identity expansion method can be extended to find cubes ( $x^3 + \frac{1}{x^3}$ ) or higher powers using $k^3 - 3k$ .

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