Question

If x-1/x=6 , find the value of x+1/x

Answer 1

Answer:

x+1/x=9 equation(1)

On the square of both sides

(x+1/x)^2= x^2 +( 1/x) ^2 +2. equation (2)

Putting the value of equation 1 into equation 2

(9)^2=x^2+(1/x)^2+2

x^2+(1/x)^2=81–2

x^2+( 1/x)^2=79 equation (3)

(x-1/x)^2= x^2+(1/x)^2–2 equation (4)

Putting the value of equation 3 into equation 4

(x-1/x)^2=79–2

(x-1/x)^2=77

x-1/x=√77

Answer 2

Given :

$\sf \: x \:-\: \dfrac{1}{x} \: = \: 6$

To find :

$\sf \: x \: + \: \dfrac{1}{x}$

Identity used :

• (a+b)² = a² + b² + 2ab
• (a-b)² = a² + b² - 2ab

Solution :

We know that

$\sf \: \: x \:-\: \dfrac{1}{x} \: = 6$

On Squaring both side , we get

$\sf \implies \: \bigg[ \: x \:-\: \dfrac{1}{x} \: \bigg]^2 \: = \: 6^2 \\ \\ \\ \sf \implies \: \bigg[ \: x^{2} \: + \bigg(\dfrac{1}{ {x} } \bigg) ^{2} \: - \: \bigg( 2 \times x \times \dfrac{1}{x} \bigg) \: \bigg] \: = \: 36\\ \\ \\ \sf \implies \: \bigg[ \: x^{2} \: + \dfrac{1}{ {x}^{2}} \: - 2 \: \bigg] \: = \: 36\\ \\ \sf \implies \: \: x^{2} \: + \dfrac{1}{ {x}^{2}} \: = 36 + 2 \\ \\ \sf \implies \: \: x^{2} \: + \dfrac{1}{ {x}^{2}} \: = 38 \: \: \: ........equation1$

Now we need to find x + (1/x)

$\sf \implies \: \: \bigg( \: x \: + \dfrac{1}{x} \bigg)^{2} \: = \: \bigg[\: x^{2} \: + \: \bigg(\dfrac{1}{x} \bigg)^{2} + \: \bigg(2 \times x \times \dfrac{1}{x} \bigg) \: \bigg] \: \: \\ \\ \sf \implies \: \bigg( \: x \: + \dfrac{1}{x} \bigg)^{2} \: = \: \: x^{2} \: + \: \dfrac{1}{x^{2}} \: + \: 2 \: \\ \\ \sf \: on \: putting \: value \: of \: x^{2} \: + \: \dfrac{1}{x^{2}} \: from \: equation \: 1 \: we \: get \: \\ \\ \sf \implies \: \: \bigg( \: x \: + \dfrac{1}{x} \bigg)^{2} \: = 38 \: + \: 2 \\ \\ \sf \implies \: \: \bigg( \: x \: + \dfrac{1}{x} \bigg)^{2} \: = \: 40 \\ \\ \sf \implies \: \: \bigg( \: x \: + \dfrac{1}{x} \bigg) = \sqrt{40} \\ \\ \sf \implies \: \: \bigg( \: x \: + \dfrac{1}{x} \bigg) \: = \: 2 \sqrt{10} \\ \\ \sf \implies \: \: \bigg( \: x \: + \dfrac{1}{x} \bigg) \: = \: 6.32$

ANSWER :

$\sf \bold{ \: \: \bigg( \: x \: + \dfrac{1}{x} \bigg) \: = \: \sqrt{40} \: = \: 2 \sqrt{10} \: \: = 6.32}$