Math, asked by maskedgirl04, 4 months ago

pleas help me with this​

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Answers

Answered by VishnuPriya2801
7

Question:-

If  \sf x = \dfrac{3 - \sqrt{13}}{2} , what is the value of x² + 1/x² ?

Answer:-

Given:

 \sf \:  x = \dfrac{3 -  \sqrt{13} }{2}

We have to find:

x² + 1/x²

We know that,

  • (a + b)² - 2ab = a² + b²

So,

 \implies \sf \bigg( x + \dfrac{1}{x} \bigg) ^{2}   - 2 \times  x \times \dfrac{1}{x}  =  {x}^{2}  +  \dfrac{1}{ {x}^{2} }  \\  \\  \\ \implies \sf \bigg( \frac{3 -  \sqrt{13} }{2}  +  \frac{1}{ \frac{3 -  \sqrt{13} }{2} }  \bigg) ^{2}  - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\ \implies \sf \bigg( \frac{3 -  \sqrt{13} }{2}  +  \frac{2}{3 -  \sqrt{13} }  \bigg) ^{2}  - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\ \implies \sf \bigg( \frac{(3 -  \sqrt{13} ) ^{2}  + 4}{2(3 -  \sqrt{13} )}   \bigg)^{2}   - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }

  • (a - b)² = a² + b² - 2ab

 \implies \sf \:  \bigg( \dfrac{ {3}^{2}  + ( \sqrt{13}) ^{2}  - 2(3)( \sqrt{13} ) + 4 }{2(3 -  \sqrt{13} )}  \bigg) ^{2}  - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\ \implies \sf \:  \bigg( \frac{9 + 13 - 6 \sqrt{13}  + 4}{2(3 -  \sqrt{13} )} \bigg) ^{2}   - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\ \implies \sf \:  \bigg( \frac{26 - 6 \sqrt{13} }{2(3 -  \sqrt{13} )}  \bigg) ^{2}  - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\  \implies \sf \:  \bigg( \frac{2(13 - 3 \sqrt{13} )}{2(3 -  \sqrt{13}) }  \bigg) ^{2}  - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\ \implies \sf \:  \bigg(  \frac{ - \sqrt{13} (3 -  \sqrt{13} )}{ 3 -  \sqrt{13} }  \bigg) ^{2}  - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\ \implies \sf \: {( -  \sqrt{13} )}^{2}  - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\ \sf \implies 13 - 2 =  {x}^{2}  +  \frac{1}{ {x}^{2} }  \\  \\  \\  \implies \boxed{ \sf 11 =  {x}^{2}  +  \frac{1}{ {x}^{2} }}

Answered by Anonymous
0

Question:-

If  \sf x = \dfrac{3 - \sqrt{13}}{2} , what is the value of x² + 1/x² ?

Answer:-

Given:

\sf \: x = \dfrac{3 - \sqrt{13} }{2}

We have to find:

x² + 1/x²

We know that,

(a + b)² - 2ab = a² + b²

So,

\begin{gathered} \implies \sf \bigg( x + \dfrac{1}{x} \bigg) ^{2} - 2 \times x \times \dfrac{1}{x} = {x}^{2} + \dfrac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \bigg( \frac{3 - \sqrt{13} }{2} + \frac{1}{ \frac{3 - \sqrt{13} }{2} } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \bigg( \frac{3 - \sqrt{13} }{2} + \frac{2}{3 - \sqrt{13} } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \bigg( \frac{(3 - \sqrt{13} ) ^{2} + 4}{2(3 - \sqrt{13} )} \bigg)^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \end{gathered}

(a - b)² = a² + b² - 2ab

\begin{gathered} \implies \sf \: \bigg( \dfrac{ {3}^{2} + ( \sqrt{13}) ^{2} - 2(3)( \sqrt{13} ) + 4 }{2(3 - \sqrt{13} )} \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{9 + 13 - 6 \sqrt{13} + 4}{2(3 - \sqrt{13} )} \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{26 - 6 \sqrt{13} }{2(3 - \sqrt{13} )} \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{2(13 - 3 \sqrt{13} )}{2(3 - \sqrt{13}) } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{ - \sqrt{13} (3 - \sqrt{13} )}{ 3 - \sqrt{13} } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: {( - \sqrt{13} )}^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \sf \implies 13 - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \boxed{ \sf 11 = {x}^{2} + \frac{1}{ {x}^{2} }}\end{gathered}

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