Question

If x2+1/x2=70 , find x-1/x

Answer 1

Answer:

2√17

Step-by-step explanation:

Given : ${\sf{\ \ x^2 + {\dfrac{1}{x^2}} = 70}}$

To Find : ${\sf{\ \ x - {\dfrac{1}{x}} }}$

Solution :

Squaring of ${\sf{ x - {\dfrac{1}{x}} }}$ , we get

$\Rightarrow{\sf{ \left( x - {\dfrac{1}{x}} \right) ^2}}$

${\boxed{\sf{\red{Identity \ : \ (a - b)^2 = a^2 + b^2 - 2ab}}}}$

${\sf{\red{Here, \ a = x, \ b = {\dfrac{1}{x}} }}}$

$\Rightarrow{\sf{(x)^2 + \left( {\dfrac{1}{x}} \right) ^2 - 2.x. {\dfrac{1}{x}} }}$

$\Rightarrow{\sf{x^2 + {\dfrac{1}{x^2}} - 2. {\cancel{x}} . {\dfrac{1}{ {\cancel{x}} }} }}$

$\Rightarrow{\sf{x^2 + {\dfrac{1}{x^2}} - 2 }}$

Putting known value, we get

$\Rightarrow{\sf{70 - 2}}$

$\Rightarrow{\sf{68}}$

Now, we have

${\sf{ \left( x - {\dfrac{1}{x}} \right) ^2 = 68}}$

$\Rightarrow{\sf{ x - {\dfrac{1}{x}} = {\sqrt{68}} }}$

$\Rightarrow{\sf{x - {\dfrac{1}{x}} = {\sqrt{2 \times 2 \times 17}} }}$

$\Rightarrow{\boxed{\sf{\green{x - {\dfrac{1}{x}} = 2 {\sqrt{17}} }}}}$

Answer 2

$\bf Given:\ \sf x^2\ +\ \dfrac{1}{x^2}\ =\ 70.\\\\\\\bf To\ find:\ \sf x\ -\ \dfrac{1}{x}.\\\\\\\bf Answer:\\\\\\\sf We\ know\ that\ (a\ -\ b)^2\ =\ a^2\ -\ 2ab\ +\ b^2.\\\\\\Supposing\ from\ the\ question,\ a\ =\ x\ and\ b\ =\ \dfrac{1}{x}.\\\\\\\bigg(x\ -\ \dfrac{1}{x}\bigg)^2\ =\ (x)^2\ +\ \bigg(\dfrac{1}{x}\bigg)^2\ -\ 2\ \times\ x\ \times\ \dfrac{1}{x}\\\\\\\bigg(x\ -\ \dfrac{1}{x}\bigg)^2\ =\ 70\ -\ 2\ \ \ \ \ \ \ \ \bigg[Given\ that\ x^2\ +\ \dfrac{1}{x^2}\ =\ 70.\bigg]$

$\sf \bigg(x\ -\ \dfrac{1}{x}\bigg)^2\ =\ 68\\\\\\x\ -\ \dfrac{1}{x}\ =\ \sqrt{68}$