Question

If x=(3+2√2) find x^+1/x^2

Answer 1

Solution :-

→ x = (3 + 2√2)

→ 1/x = 1/(3 + 2√2)

Rationlizing RHS part ,

→ 1/x = 1/(3 + 2√2) * (3 - 2√2) / (3 - 2√2)

using (a + b)(a - b) = a² - b² Now,

→ 1/x = (3 - 2√2) / {(3)² - (2√2)²}

→ 1/x = (3 - 2√2) / (9 - 8)

→ 1/x = (3 - 2√2)

So ,

→ (x + 1/x) = (3 + 2√2) + (3 - 2√2)

→ (x + 1/x) = 6

squaring both sides Now,

→ (x + 1/x)² = 6²

using (a + b)² = a² + b² + 2ab in LHS now,

→ x² + 1/x² + 2 * x * 1/x = 36

→ x² + 1/x² + 2 = 36

→ (x² + 1/x²) = 36 - 2

→ (x² + 1/x²) = 34 (Ans.)

Answer 2

➝ Question :

If $x = 3 + 2\sqrt{2}$,then find the value of $x^{2} + \dfrac{1}{x^{2}}$

➝ To Find :

The value of the equation :p

$\boxed{x^{2} + \dfrac{1}{x^{2}}}$

➝ Given :

The value of x....

• $x = 3 + 2\sqrt{2}$[Equation...(ii)]p

➝ We Know :

• $\mathtt{(a + b)^{2} = a^{2} + b^{2} + 2ab}$

• $\mathtt{a^{2} - b^{2} = (a + b)(a - b)}$

➝ Concept :

We know that ,if $x = 3 + 2\sqrt{2}$ ,then $\dfrac{1}{x} = 3 - 2\sqrt{2}$

Proof :

If $x = 3 + 2\sqrt{2}$

then,

$\dfrac{1}{x} = \dfrac{1}{3 + 2\sqrt{2}}$

$\Rightarrow \dfrac{1}{x} = 3 + 2\sqrt{2} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}}$

$\Rightarrow \dfrac{1}{x} = \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2} \times 3 - 2\sqrt{2}}$

Using the identity :

$a^{2} - b^{2} = (a + b)(a - b)$

we get :

$\Rightarrow \dfrac{1}{x} = \dfrac{3 - 2\sqrt{2}}{3^{2} - \big(2\sqrt{2}\big)^{2}}$

$\Rightarrow \dfrac{1}{x} = \dfrac{3 - 2\sqrt{2}}{9 - 8}$

$\Rightarrow \dfrac{1}{x} = \dfrac{3 - 2\sqrt{2}}{1}$

$\Rightarrow \dfrac{1}{x} = 3 - 2\sqrt{2}$

So ,we get the value of $\dfrac{1}{x}$ as $3 - 2\sqrt{2}$[Equation...(ii)]

➝ Solution :

Equation...(i)

$\mathtt{x = 3 + 2\sqrt{2}}$

Equation...(ii)

$\mathtt{3 - 2\sqrt{2}}$

On adding equation (i) and (ii) ,we get :-

ATP :

$\boxed{x + \dfrac{1}{x}}$

$\mathtt{\Rightarrow x + \dfrac{1}{x}}$

$\mathtt{\Rightarrow 3 + 2\sqrt{2} + 3 - 2\sqrt{2}}$

$\mathtt{\Rightarrow 3 + \cancel{2\sqrt{2}} + 3 - \cancel{2\sqrt{2}}}$

$\mathtt{\Rightarrow 3 + 3}$

$\mathtt{\Rightarrow 6}$

Hence ,

$\boxed{\therefore x + \dfrac{1}{x} = 6}$

Given Equation :

$\mathtt{x + \dfrac{1}{x} = 6}$

Squaring on both the sides ,we get :

$\mathtt{\left(x + \dfrac{1}{x}\right)^{2} = 6^{2}}$

Using the identity :

$\mathtt{(a + b)^{2} = a^{2} + b^{2} + 2ab}$

we get ,

$\mathtt{x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} + 2 \times x \times \dfrac{1}{x} = 36}$

$\mathtt{x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} + 2 \times \cancel{x} \times \dfrac{1}{\cancel{x}} = 36}$

$\mathtt{x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} + 2 = 36}$

$\mathtt{x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} = 36 - 2}$

$\mathtt{x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} = 34}$

Hence ,the value of $x^{2} + \dfrac{1}{x^{2}}$ is 34....

➝ Extra Information :

Some useful identities :

• $(a + b)^{2} = a^{2} + b^{2} - 2ab$

• $a^{2} + b^{2} = (a + b)^{2} - 2ab$

• $\bigg(x - \dfrac{1}{x}\bigg)^{2} = x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} - 2$