Question

If x=3+ 2
then
13 - 2
find the value
of x²​

Answer 1

Step-by-step explanation:

As per the provided information in the given question, we have :

$\longmapsto\rm { x = \dfrac{\sqrt{3} +\sqrt{2}}{\sqrt{3} -\sqrt{2}} }$

We are asked to calculate the value of x². In order to calculate the value of x², we need to rationalize the denominator of the fraction.

In order to rationalise the denominator of the fraction, we multiply the rationalising factor of the denominator with both the numerator and the denominator of the fraction.

Rationalising the denominator of the fraction :

Here, denominator is in the form of (a - b), rationalising factor of (a - b) is (a + b). Hence, rationalising factor of (√3 - √2) is (√3 + √2).

Multiplying (√3 + √2) with both the numerator and the denominator of the fraction.

$\longmapsto\rm { x = \dfrac{\sqrt{3} +\sqrt{2}}{\sqrt{3} -\sqrt{2}} \times \dfrac{\sqrt{3} +\sqrt{2}}{\sqrt{3} +\sqrt{2}} }$

Rearranging the terms.

$\longmapsto\rm { x = \dfrac{(\sqrt{3} +\sqrt{2})^2}{(\sqrt{3} -\sqrt{2})(\sqrt{3} +\sqrt{2}) }}$

Now, using the identity,

• (a + b)² = a² + b² + 2ab
• (a + b)(a - b) = a² - b²

$\longmapsto\rm { x = \dfrac{(\sqrt{3})^2 +(\sqrt{2})^2 + 2\sqrt{2\times 3}}{(\sqrt{3})^2 -(\sqrt{2})^2 }}$

Now, writing the squares of the numbers.

$\longmapsto\rm { x = \dfrac{3 + 2 + 2\sqrt{6}}{3- 2}}$

Performing arithmetic operations in the numerator and the denominator.

$\longmapsto\rm { x = \dfrac{5 + 2\sqrt{6}}{1}}$

It can be written as,

$\longmapsto\rm { x = 5 + 2\sqrt{6}}$

Now, finding the value of x².

If,

$\longmapsto\rm { x = 5 + 2\sqrt{6}}$

So,

$\longmapsto\rm { x^2 = (5 + 2\sqrt{6})^2}$

We know that,

• (a + b)² = a² + b² + 2ab

$\longmapsto\rm { x^2 = (5)^2 + (2\sqrt{6})^2 + 2( 5 \times 2 \sqrt{6} )}$

Simplifying further by the writing the squares and performing multiplication.

$\longmapsto\rm { x^2 = 25 + 24 + 2( 10 \sqrt{6} )}$

Performing addition and multiplication.

$\longmapsto\bf { x^2 = 49 + 20 \sqrt{6}}$

Hence, we got the required answer!

Answer 2

$\huge\dag\sf\:\underline{\red{Question}}:$

$\longmapsto \: \sf If \: x = \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} , \textsf{Find the value of }x^2$

$\huge\dag\sf\:\underline{\red{Solution}}:$

$\bigstar\:\underline{\textbf{Concept to understand}} :$

⠀⠀⠀ ⠀ • We have value of x. Now just square it out to get the value of x² . After that we have to rationalize it to make it in simple form.

$\bigstar\:\underline{\textbf{Points to remember}} :$

⠀⠀⠀ ⠀ • a² = a × a

⠀⠀⠀ ⠀ • (a+b)² = a² + b² + 2ab

⠀⠀⠀ ⠀ • a²-b² = (a+b)(a-b)

⠀⠀⠀ ⠀ • (a-b)² = a² + b² - 2ab

⠀⠀⠀ ⠀ • $\sf\left(\dfrac{a}{b}\right)^m = \dfrac{(a)^m}{(b)^m}$

Let's start solving,

$\longmapsto \sf x = \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\\\ \sf \longmapsto x^2 = \left(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)^2 \\\\ \sf \implies x^2 = \dfrac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})^2} \\\\ \sf \implies x^2 = \dfrac{(\sqrt{3}^2+\sqrt{2}^2+2\times\sqrt{3}\times\sqrt{2})}{(\sqrt{3}^2+\sqrt{2}^2-2\times\sqrt{3}\times\sqrt{2})} \\\\ \sf \implies x^2 = \dfrac{3+2+2\sqrt{6}}{3+2-2\sqrt{6}}\\\\ \sf \implies x^2 = \dfrac{5+2\sqrt{6}}{5-2\sqrt{6}} \:\:\:\: \red\bigstar$

⠀⠀⠀ ⠀ ✰ Here we have got the value of x². Now we have to rationalize it to get it in simple form.

$\mapsto\sf\underline{Rationalising }:$

⠀⠀⠀ ⠀ • To rationalise we have to Multiply numerator and denominator by Conjugate of the denominator .

⠀⠀⠀ ⠀ • Conjugate of 5-2√6 is 5+2√6.

$\sf = \dfrac{5+2\sqrt{6}}{5-2\sqrt{6}} \times \dfrac{5+2\sqrt{6}}{5+2\sqrt{6}} \\\\\\ =\dfrac{(5+2\sqrt{6})^2}{\sqrt{5}^2 - (22\sqrt{6})^2} \\\\\\ = \dfrac{25 + 24 + 20\sqrt{6}}{25-24} \\\\\\ = \dfrac{49+ 20\sqrt{6} }{1} \\\\\\ \implies 49+ 20\sqrt{6} \\\\ ::\implies x^2 = 49+ 20\sqrt{6}$

Therefore,

$\\\;\:\:\:\red\bigstar\;\sf Value\;of\; x^2 \;is\; 49+ 20\sqrt{6}$