Math, asked by anweshadeb14, 5 hours ago

(sqrt(2) + sqrt(3))/(3sqrt(2) - 2sqrt(3)) = a - b * sqrt(6) . Find a and b.​

Answers

Answered by Yuseong
321

Given Equation :

 \sf {  \dfrac{ \sqrt{2} +  \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3}  }  = a - b \sqrt{6} }

To calculate :

Value of a and b.

Clarification :

In order to calculate the value of a and b, firstly we need to rationalize the denominator in L.H.S. After that, by comparing L.H.S and R.H.S, we'll finr the value of a and b.

How to rationalize the denominator ?

  • To rationalize the denominator, we simply multiply the numerator and the denominator with the rationalizing factor of the denominator.
  • After that, by using algebraic identities, we simplify further.

Explication of steps :

\longrightarrow \sf {  \dfrac{ \sqrt{2} +  \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3}  }  = a - b \sqrt{6} }

Rationalising the denominator in LHS :

  • Rationalizing factor of (a - b) is (a + b). So, rationalizing factor of (3√2 - 2√3) is (3√2 + 2√3).

\longrightarrow \sf {  \dfrac{ \sqrt{2} +  \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3}  }  \times   \dfrac{3 \sqrt{2}   + 2 \sqrt{3} }{3 \sqrt{2}   + 2 \sqrt{3} } }\\

\longrightarrow \sf {  \dfrac{( \sqrt{2} +  \sqrt{3} )(3 \sqrt{2}   + 2 \sqrt{3})}{(3 \sqrt{2} - 2 \sqrt{3})(3 \sqrt{2}   + 2 \sqrt{3})  } }  \\

As we know that,

  • (a - b)(a + b) = a² - b²

\longrightarrow \sf {  \dfrac{( \sqrt{2}  )(3 \sqrt{2}   + 2 \sqrt{3})+  \sqrt{3}(3 \sqrt{2}   + 2 \sqrt{3})}{ {(3   \sqrt{ 2}) }^{2} -  { ( 2   \sqrt{3}) }^{2}   } }  \\

\longrightarrow \sf {   \dfrac{6 + 2 \sqrt{ 6} + 3 \sqrt{6}  + 6 }{18 - 12}  }  \\

\longrightarrow \sf {   \dfrac{12 + (2  + 3)\sqrt{ 6}  }{6}  }  \\

\longrightarrow \sf {  2  +  \dfrac{5 \sqrt{6} }{6}  }  \\

\longrightarrow  \underline{\boxed{\sf {  2  +  \dfrac{5}{6}  \sqrt{6} } }}  \\

Now, in RHS we have :

 \longrightarrow \sf {  a - b \sqrt{6} }

From, LHS and RHS, we have :

\longrightarrow  \sf {  2  +  \dfrac{5}{6}  \sqrt{6} \Leftrightarrow a - b \sqrt{6} }    \\

\longrightarrow  \underline{\boxed{ {   \begin{array}{cc}  \sf{a = 2 }  \\  \\ \sf b =  -  \dfrac{5    }{6} \end{array}} }} \:  \bigstar  \\

Answered by js9116294
28

Step-by-step explanation:

See this Sum carefully Bro

Attachments:
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