Question

ch-1 class 9 maths response only if u know​

Answer 1

Given Question :-

$\sf \: If \: x = \dfrac{1}{2 - \sqrt{3} }, \: find \: the \: value \: of \: {x}^{2} - 4x + 1$

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

Given that,

$\rm :\longmapsto\:x = \dfrac{1}{2 - \sqrt{3} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:x = \dfrac{1}{2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 + \sqrt{3} }$

We know

$\red{\boxed{ \tt{ \: (x - y)(x + y) = {x}^{2} - {y}^{2} \: }}}$

So, using this identity, we get

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

$\rm :\longmapsto\:x = \dfrac{2 + \sqrt{3} }{ 4 - 3 }$

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

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

can be rewritten as

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

On squaring both sides, we get

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

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

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

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

$\rm :\longmapsto\: {x}^{2} - 4x + 4 - 3 = 0$

$\rm :\longmapsto\: {x}^{2} - 4x + 1= 0$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {x}^{2} - 4x + 1 = 0 \: }}}$

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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)