Math, asked by abhinandangoel005, 11 months ago

x^2+1/x^2=7, find 2x^2-2/x^2

Answers

Answered by aami21

${\red{\bold{\huge{\mathfrak{Answer}}}}}$

${x}^{2} + 1 = 7 {x}^{2} \\ 1 = 6 {x}^{2} \\ {x}^{2} = \frac{1}{6}$

substitute it

$\frac{2 {x}^{2} - 2}{2 {x}^{2} } = \frac{2 \frac{1}{6} - 2 }{2 \times \frac{1}{6} } \\ \frac{ \frac{1}{3} - 2 }{ \frac{1}{3} } = \frac{ \frac{ - 5}{3} }{ \frac{1}{3} } = - 5$

${\boxed{\blue{\huge{\bold{-5}}}}}$

abhinandangoel005: I am a student it is my assignment

aami21: even i am a student

abhinandangoel005: You are Senior

aami21: so wt?

aami21: then post your ans with steps

aami21: thnx

Answered by abdul9838

$<b> <body bgcolor = "skyblue">$

$\small \bf \red{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \red{ \huge \: solution} \\ \\ \small \bf \red{ \huge \: ans \: is \: 6\sqrt{5} } \\ \\ \small \bf \red{ \underline{ \underline{step \: by \: step \: explaination}}} \\ \\ \small \bf \red{given \: that} \\ \\ \small \bf \red{ {x}^{2} + \frac{1}{ {x}^{2} } = 7 } \\ \\ \small \bf \red{ \underline{as \: we \: can \: write}} \\ \\ \small \bf \red{ {x}^{2} + \frac{1}{ {x}^{2} } = (x + \frac{1}{x} )^{2} - 2 \times x \times \frac{1}{x} } \\ \\ \\ \\ \small \bf \red{here \: it \: does \: not \: matter \: what \: is \: the \: } \\ \small \bf \red{power \: of \: x} \\ \\ \small \bf \red{ \huge \: so} \\ \\ \small \bf \red{ \underline{as \: we \: can \: wite}} \\ \\ \\ \\ \small \bf \red{ {x}^{n} + \frac{1}{ {x}^{n} } = k } \\ \\ \small \bf \red{and \: it \: is \: equall \: } \\ \\ \small \bf \red{(x + \frac{1}{x} )^{2} = k - 2 } \\ \\ \small \bf \red{but \: here \: k = 7 \: ( \underline{according \: to \: question)}} \\ \\ \small \bf \red{now} \\ \\ \small \bf \red{(x - \frac{1}{x})^{2} + 2 \times x \times \frac{1}{x} = {x}^{2} - \frac{1}{ {x}^{2} } } \\ \\ \small \bf \red{using \: this \: formula} \\ \\ \small \bf \red{ {x}^{2} - \frac{1}{ {x}^{2} } = \sqrt{{k}^{2} - 4 } } \\ \\ \small \bf \red{ { {x}^{2} - \frac{1}{ {x}^{2} } = \sqrt{ {7}^{2} - 4 } }} \\ \\ \small \bf \red{ {x}^{2} - \frac{1}{ {x}^{2} } = \sqrt{49 - 4} } \\ \\ \small \bf \red{ {x}^{2} - \frac{1}{ {x}^{2} } = \sqrt{45} } \\ \\ \small \bf \red{ {x}^{2} - \frac{1}{ {x}^{2} } = 3 \sqrt{5} } \\ \\ \small \bf \red{here \: the \: value \: of} \\ \\ \small \bf \red{ {x}^{2} - \frac{1}{ {x}^{2} } = 3 \sqrt{5} } \\ \\ \small \bf \red{but} \\ \\ \small \bf \red{cofficient \: of \: {x}^{2} = 2} \\ \\ \small \bf \red{ \huge \: so} \\ \\ \small \bf \red{2 \times 3 \sqrt{5} = 6 \sqrt{5} } \\ \\ \small \bf \red{ \huge \: ans \: is \: 6 \sqrt{5} }$

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