Question

if
$x + \frac{1}{x} = 2 \:$
then prove that
${x}^{8} - 2 {x}^{4} + 1 = 0$

​

Answer 1

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

$\: \: \: \: \: \: \: \: \bull \sf \: \: \: x + \dfrac{1}{x} = 2$

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

$\: \: \: \: \: \: \: \: \bull \sf \: \: \: {x}^{8} - {2x}^{4} + 1 = 0$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

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

Given that,

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

On squaring both sides, we get

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

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times \cancel x \times \dfrac{1}{\cancel{x}} = 4$

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

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

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

On squaring both sides, we get

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

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 \times \cancel {x}^{2} \times \dfrac{1}{\cancel {x}^{2} } = 4$

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

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

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

$\rm :\longmapsto\:\dfrac{ {x}^{8} + 1}{ {x}^{4} } = 2$

$\rm :\longmapsto\: {x}^{8} + 1 = {2x}^{4}$

$\bf\implies \: {x}^{8} - 2 {x}^{4} + 1 = 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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)