Question

If x= 1/1+√2 then value of x² + 2x + 3 is​

Answer 1

$\large\underline{\sf{Given- }}$

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:The \: value \: of \: {x}^{2} + 2x + 3$

$\large\underline{\sf{Solution-}}$

Given that

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

☆ On rationalizing the denominator, we get

$\rm :\longmapsto\:x = \dfrac{1}{ \sqrt{2} + 1 } \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1 }$

$\rm :\longmapsto\:x = \: \dfrac{ \sqrt{2} - 1 }{ {( \sqrt{2} )}^{2} - {(1)}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm :\longmapsto\:x = \dfrac{ \sqrt{2} - 1 }{2 - 1}$

$\rm :\implies\:x = \sqrt{2} - 1$

☆ can be rewritten as

$\rm :\implies\:x + 1= \sqrt{2}$

☆ On squaring both sides, we get

$\rm :\longmapsto\: {(x + 1)}^{2} = {( \sqrt{2})}^{2}$

$\rm :\longmapsto\: {x}^{2} + 1 + 2x = 2$

☆ Adding 2 on both sides, we get

$\rm :\longmapsto\: {x}^{2} + 1 + 2x + 2 = 2 + 2$

$\rm :\longmapsto\: {x}^{2} + 2x + 3 =4$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \underbrace{ \large{ \boxed{ \sf \: {x}^{2} + 2x + 3 = 4}}}}$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

Hey mate your answer is spotted in the upper image ⬆️⬆️⬆️⬆️⬆️

Hope that my answer satisfied you

and helps you

THNXS