Answer:
[tex]\boxed{\bf\:x - \dfrac{1}{x} = \pm \: 2 \: }\\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 6 \\ [/tex]
On Subtracting 2 from both sides, we get
[tex]\sf\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 6 - 2 \\ [/tex]
[tex]\sf\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times x \times \dfrac{1}{x} = 4 \\ [/tex]
[tex]\sf\: {\bigg( x - \dfrac{1}{x} \bigg) }^{2} = 4 \\ [/tex]
[tex]\sf\: {\bigg( x - \dfrac{1}{x} \bigg) }^{2} = {(2)}^{2} \\ [/tex]
[tex]\implies\sf\:x - \dfrac{1}{x} = \pm \: 2 \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:x - \dfrac{1}{x} = \pm \: 2 \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
[tex](x - \frac{1}{x} ) {}^{2} \\ \\ = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times x \times \frac{1}{x} \\ \\ = {x}^{2} + \frac{1}{ {x}^{2} } - 2[/tex]
[tex](x - \frac{1}{x} ) {}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ (x - \frac{1}{x} ) {}^{2} = 6 - 2 \\ \\ (x - \frac{1}{x} ) {}^{2} = 4 \\ \\ x - \frac{1}{x} = \sqrt{4} \\ \\ x - \frac{1}{x} = 2[/tex]
Answer:
[tex]\boxed{\bf\:x - \dfrac{1}{x} = \pm \: 2 \: }\\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 6 \\ [/tex]
On Subtracting 2 from both sides, we get
[tex]\sf\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 6 - 2 \\ [/tex]
[tex]\sf\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times x \times \dfrac{1}{x} = 4 \\ [/tex]
[tex]\sf\: {\bigg( x - \dfrac{1}{x} \bigg) }^{2} = 4 \\ [/tex]
[tex]\sf\: {\bigg( x - \dfrac{1}{x} \bigg) }^{2} = {(2)}^{2} \\ [/tex]
[tex]\implies\sf\:x - \dfrac{1}{x} = \pm \: 2 \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:x - \dfrac{1}{x} = \pm \: 2 \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]