Question

If x-1/x =-2 then find the value of x+1/x

Answer 1

HEY MATE HERE IS YOUR ANSWER......

Answer 2

$\underline{\textsf{Question :}}$

$\textsf{If}\:\bold{x-\frac{1}{x} =-2}$

$\textsf{then}\:x +\frac{1}{x} =?$

$\underline{\textsf{Solution :}}$

$\textsf{Given,}\:\:\bold{x-\frac{1}{x} =-2}$

$\underline{\textsf{Squaring both side we get}}$

$\implies \bold{x^{2}+\frac{1}{x^{2}} -2*x*\frac{1}{x} =(-2)^{2}}$

$\implies \bold{x^{2} +\frac{1}{x^{2} } -2=4}}$

$\implies \bold{x^{2} +\frac{1}{x^{2} } =4+2}$

$\implies \boxed{\bold{x^{2} +\frac{1}{x^{2}} =6}}$

$\bold{x^{2}+\dfrac{1}{x^{2} }=6 }\\\\\\\implies{\bold{(x+\dfrac{1}{x}){^2}-2*x*\frac{1}{x} =6}}\\\\\\\implies{\bold{(x+\dfrac{1}{x})^{2}=6+2=8}}\\\\\\\implies{\bold{x+\dfrac{1}{x}=\sqrt{8} =2\sqrt{2}. }}\\\\\\\boxed{\boxed{\bold{x+\dfrac{1}{x}=2\sqrt{2}. }}}$

$\underline{\textsf{Identity Rules :}}$

$1.\:\bold{(a+b)^{2}=a^{2}+2ab+b^{2}}$

$2.\:\bold{(a-b)^{2}=a^{2}-2ab+b^{2}}$

$3.\:\bold{(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}}$

$4.\:\bold{(a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}}$

$5.\:\bold{a^{2}+b^{2}=(a+b)^{2}-2ab}$

$6.\:\bold{a^{2}-b^{2}=(a+b)(a-b)}$