Question

Can someone please explain me how do we solve this sum.​

Answer 1

Answer :-

Given :-

• $\sf x + \dfrac{1}{x} = 3$

To Find :-

• $\sf x^2 + \dfrac{1}{x^2}$

Solution :-

$\implies\sf x + \dfrac{1}{x} = 3$

Squaring on both side :-

$\implies\sf (x + \dfrac{1}{x})^2 = 3^2$

• ( a + b )² = a² + b² + 2ab

$\implies\sf x^2 + \dfrac{1}{x^2} + 2 \times \not x \times \dfrac{1}{\not x} = 9$

$\implies\sf x^2 + \dfrac{1}{x^2} + 2 = 9$

$\implies\sf x^2 + \dfrac{1}{x^2} = 9-2$

$\implies\sf x^2 + \dfrac{1}{x^2} = 7$

$\boxed{\bf Value \: of \: x^2 + \dfrac{1}{x^2} = 7}$

Additional information :-

$\boxed{\bigstar\:\:\textbf{\textsf{Algebraic\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})$

Answer 2

We know that:

$x \times \frac{1}{x} = 3$

Squaring both sides

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

x²+1/x²+2=9x²+1/x²+2=9

x²+1/x²=7x²+1/x²=7