Answer:
a=5
b=2
Step-by-step explanation:
√3+√2/√3-√2=(a+b√6)
(√3+√2)(√3+√2)/(√3-√2)(√3+√2)=(a+b√6) [: Rationalizing]
3+2 +2√6/1 = a+b√6
5 + 2√6 = a+b√6
[tex]\qquad\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: a=5 \qquad \: \\ \\& \qquad \:\sf \: b=2\end{aligned}} \qquad \: \\ \\ [/tex]
Given that,
[tex]\sf \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6} \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = a + b \sqrt{6} \\ \\ [/tex]
[tex]\sf \: \dfrac{ (\sqrt{3} + \sqrt{2})^{2} }{( \sqrt{3})^{2} - (\sqrt{2})^{2} } = a + b \sqrt{6} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: \dfrac{ (\sqrt{3})^{2} + (\sqrt{2})^{2} + 2( \sqrt{3} )( \sqrt{2} )}{3 - 2} = a + b \sqrt{6} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }} \\ \\ [/tex]
[tex]\sf \: \dfrac{3 + 2 + 2 \sqrt{6} }{1} = a + b \sqrt{6} \\ \\ [/tex]
[tex]\sf \: 5 + 2 \sqrt{6} = a + b \sqrt{6} \\ \\ [/tex]
On comparing, we get
[tex]\sf\implies \: a = 5 \: \: and \: \: b = 2 \\ \\ [/tex]
Hence,
[tex]\qquad \qquad\:\boxed{\begin{aligned}& \qquad \:\sf \: a=5 \qquad \: \\ \\& \qquad \:\sf \: b=2\end{aligned}} \qquad \: \\ \\ [/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]
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Answers & Comments
Answer:
a=5
b=2
Step-by-step explanation:
√3+√2/√3-√2=(a+b√6)
(√3+√2)(√3+√2)/(√3-√2)(√3+√2)=(a+b√6) [: Rationalizing]
3+2 +2√6/1 = a+b√6
5 + 2√6 = a+b√6
Verified answer
Answer:
[tex]\qquad\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: a=5 \qquad \: \\ \\& \qquad \:\sf \: b=2\end{aligned}} \qquad \: \\ \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6} \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = a + b \sqrt{6} \\ \\ [/tex]
[tex]\sf \: \dfrac{ (\sqrt{3} + \sqrt{2})^{2} }{( \sqrt{3})^{2} - (\sqrt{2})^{2} } = a + b \sqrt{6} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: \dfrac{ (\sqrt{3})^{2} + (\sqrt{2})^{2} + 2( \sqrt{3} )( \sqrt{2} )}{3 - 2} = a + b \sqrt{6} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }} \\ \\ [/tex]
[tex]\sf \: \dfrac{3 + 2 + 2 \sqrt{6} }{1} = a + b \sqrt{6} \\ \\ [/tex]
[tex]\sf \: 5 + 2 \sqrt{6} = a + b \sqrt{6} \\ \\ [/tex]
On comparing, we get
[tex]\sf\implies \: a = 5 \: \: and \: \: b = 2 \\ \\ [/tex]
Hence,
[tex]\qquad \qquad\:\boxed{\begin{aligned}& \qquad \:\sf \: a=5 \qquad \: \\ \\& \qquad \:\sf \: b=2\end{aligned}} \qquad \: \\ \\ [/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]