Question

If d-1/d=6, evaluate (1) d + 1/d (2)d²+1/d² and(3) d⁴+1/d⁴​

Answer 1

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

Given that,

$\rm :\longmapsto\:d - \dfrac{1}{d} = 6$

We know that,

$\rm :\longmapsto\: {\bigg[d + \dfrac{1}{d} \bigg]}^{2} - {\bigg[d - \dfrac{1}{d} \bigg]}^{2} = 4$

$\rm :\longmapsto\: {\bigg[d + \dfrac{1}{d} \bigg]}^{2} - {\bigg[6 \bigg]}^{2} = 4$

$\rm :\longmapsto\: {\bigg[d + \dfrac{1}{d} \bigg]}^{2} - 36 = 4$

$\rm :\longmapsto\: {\bigg[d + \dfrac{1}{d} \bigg]}^{2} = 4 + 36$

$\rm :\longmapsto\: {\bigg[d + \dfrac{1}{d} \bigg]}^{2} = 40$

$\rm :\longmapsto\:d + \dfrac{1}{d} = \: \pm \: \sqrt{40} = \: \pm \: 2 \sqrt{10}$

Given that,

$\rm :\longmapsto\:d - \dfrac{1}{d} = 6$

On squaring both sides, we get

$\rm :\longmapsto\: {\bigg[d - \dfrac{1}{d} \bigg]}^{2} = {6}^{2}$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\: {d}^{2} + \dfrac{1}{ {d}^{2} } - 2 \times d \times \dfrac{1}{d} = 36$

$\rm :\longmapsto\: {d}^{2} + \dfrac{1}{ {d}^{2} } - 2 = 36$

$\rm :\longmapsto\: {d}^{2} + \dfrac{1}{ {d}^{2} } = 36 + 2$

$\red{\bf\implies \: \boxed{ \sf{ \:{d}^{2} + \dfrac{1}{ {d}^{2} } = 38}}}$

On squaring both sides, we get

$\rm :\longmapsto\: {\bigg[ {d}^{2} + \dfrac{1}{ {d}^{2} } \bigg]}^{2} = {38}^{2}$

$\rm :\longmapsto\: {d}^{4} + \dfrac{1}{ {d}^{4} } + 2 \times {d}^{2} \times \dfrac{1}{ {d}^{2} } = 1444$

$\rm :\longmapsto\: {d}^{4} + \dfrac{1}{ {d}^{4} } + 2 = 1444$

$\rm :\longmapsto\: {d}^{4} + \dfrac{1}{ {d}^{4} }= 1444 - 2$

$\red{ \bf\implies \:\boxed{ \sf{ \: {d}^{4} + \dfrac{1}{ {d}^{4} }= 1442}}}$

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)