Question

$is \: it \: correct.$
Is it correct?

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Solve it if it is wrong!( No information )!!



I have tried to solve it but I am freezed since, I got it first time!​

Answer 1

No the answer is wrong.

Given Data :

$\rm \: If \: x - \frac{1}{x} =\:\frac{1}{6}$

To find :

${x}^{2} + \frac{1}{x^{2} }$

Formula Used :

(a–b)² = a²+b² – 2ab

Solution :

$(x - \frac{1}{x} )^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times x \times \frac{1}{x}$

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

[tex]\rm \:So\:we\:need\:to\:put\:the\:value \: 1/6 \: in \: place \: of \: {(x - \frac{1}{x}) }^{2}[/tex]

${(</u></strong><strong><u>\frac{1}{6}</u></strong><strong><u>)}^{2}={x}^{2}+\frac{1}{ {x}^{2} }</u></strong><strong><u>-</u></strong><strong><u>2$

${x}^{2} + \frac{1}{ {x}^{2} }</u></strong><strong><u>=</u></strong><strong><u>\frac{1}{</u></strong><strong><u>3</u></strong><strong><u>6</u></strong><strong><u>}</u></strong><strong><u> - 2$

${x}^{2} + \frac{1}{ {x}^{2} } =</u></strong><strong><u>\frac{</u></strong><strong><u>–</u></strong><strong><u>7</u></strong><strong><u>1</u></strong><strong><u>}{</u></strong><strong><u>3</u></strong><strong><u>6</u></strong><strong><u>}</u></strong><strong><u>$

[tex]Correct\:answer=\frac{–71}{36}[/tex]

Answer 2

$\huge \bigstar { \underline{ \mathtt{ \red{A}{ \pink{n}{ \color{blue}{s}{ \color{gold}{w}{ \color{aqua}{e}{ \color{lime}{r}}}}}}}}}★$

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