If x = 2 + sqrt(3) and y = 2 - sqrt(3) then value of x2 + y2 =

Option a: 14. Solution: We are given: x = 2 + sqrt(3) y = 2 - sqrt(3) Let us first find the sum and the product of x and y : 1) Sum of x and y : x + y = ( 2 + sqrt(3) ) + ( 2 - sqrt(3) ) = 2 + 2 = 4 2) Product of x and y (using the difference of squares identity, (A+B)(A-B) = A2 - B2 ): xy = ( 2 + sqrt(3) ) ( 2 - sqrt(3) ) = 22 - ( sqrt(3) )2 = 4 - 3 = 1 Now, we use the algebraic identity for the sum of squares: x2 + y2 = (x+y)2 - 2xy Substituting the values of (x+y) and xy : x2 + y2 = (4)2 - 2(1) = 16 - 2 = 14 Hence, Option A is the correct answer.

If $x = 2 + \sqrt{3}$ and $y = 2 - \sqrt{3}$ then value of $x^2 + y^2 =$

The correct answer is Option a: 14. We are given: $$x = 2 + \sqrt{3}$$ $$y = 2 - \sqrt{3}$$ Let us first find the sum and the product of $x$ and $y$: 1) Sum of $x$ and $y$: $$x + y = \left( 2 + \sqrt{3} \right) + \left( 2 - \sqrt{3} \right) = 2 + 2 = 4$$ 2) Product of $x$ and $y$ (using the difference of squares identity, $(A+B)(A-B) = A^2 - B^2$): $$xy = \left( 2 + \sqrt{3} \right) \left( 2 - \sqrt{3} \right) = 2^2 - \left( \sqrt{3} \right)^2 = 4 - 3 = 1$$ Now, we use the algebraic identity for the sum of squares: $$x^2 + y^2 = (x+y)^2 - 2xy$$ Substituting the values of $(x+y)$ and $xy$: $$x^2 + y^2 = (4)^2 - 2(1) = 16 - 2 = 14$$ Hence, **Option A** is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQMTP Dec 23 Series IIQuestion 923 of 305

All Questions

If $\displaystyle x = 2 + \sqrt{3}$ and $\displaystyle y = 2 - \sqrt{3}$ then value of $\displaystyle x^2 + y^2 =$

Options

A14

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Correct Answer

✅ Option a — 14

All Options:

Detailed Solution & Explanation

We are given:

x = 2 + \sqrt{3}

y = 2 - \sqrt{3}

Let us first find the sum and the product of

\displaystyle x

and

\displaystyle y

:
1) Sum of

\displaystyle x

and

\displaystyle y

x + y = \left( 2 + \sqrt{3} \right) + \left( 2 - \sqrt{3} \right) = 2 + 2 = 4

2) Product of

\displaystyle x

and

\displaystyle y

(using the difference of squares identity,

\displaystyle (A+B)(A-B) = A^2 - B^2

xy = \left( 2 + \sqrt{3} \right) \left( 2 - \sqrt{3} \right) = 2^2 - \left( \sqrt{3} \right)^2 = 4 - 3 = 1

Now, we use the algebraic identity for the sum of squares:

x^2 + y^2 = (x+y)^2 - 2xy

Substituting the values of

\displaystyle (x+y)

and

\displaystyle xy

x^2 + y^2 = (4)^2 - 2(1) = 16 - 2 = 14

Hence, **Option A** is the correct answer.

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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If x=2+3\displaystyle x = 2 + \sqrt{3}x=2+3​ and y=2−3\displaystyle y = 2 - \sqrt{3}y=2−3​ then value of x2+y2=\displaystyle x^2 + y^2 =x2+y2=