Math, asked by shaileah1234, 1 year ago

if ✓3+✓2/✓3-✓2=a+b✓6, find the value of a and b​

Answers

Answered by ihrishi
1

Answer:

 \frac{ \sqrt{3} +  \sqrt{2}  }{ \sqrt{3}  -  \sqrt{2} }  = a + b \sqrt{6}  \\ \frac{ \sqrt{3} +  \sqrt{2}  }{ \sqrt{3}  -  \sqrt{2} }  \times  \frac{\sqrt{3} +  \sqrt{2} }{\sqrt{3} +  \sqrt{2} }  = a + b \sqrt{6}  \\   \frac{( \sqrt{3 }  +  \sqrt{2})^{2}}{  {( \sqrt{3}) }^{2}  -  {( \sqrt{2} )}^{2}  }  = a + b \sqrt{6}  \\  \frac{( \sqrt{3})^{2}   +  {( \sqrt{2} )}^{2} + 2 \sqrt{3}   \sqrt{2} }{3 - 2} = a + b \sqrt{6}  \\  \frac{3 + 2 + 2 \sqrt{6} }{1} = a + b \sqrt{6}  \\ 5 + 2 \sqrt{6}  = a + b \sqrt{6} \\  \implies \: a = 5  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \: b = 2

Answered by MavisRee
1

Answer:

Value of a is 5 and value of b is 2

Step-by-step explanation:

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

Consider LHS:

\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}

Applying componendo and dividendo theorem, we get

\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\\\\\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}\\\\ \frac{(\,\sqrt{3} + \sqrt{2})\,(\,\sqrt{3} + \sqrt{2})\,}{(\,\sqrt{3} - \sqrt{2})\,(\,\sqrt{3} + \sqrt{2})\,}\\\\ \frac{(\,\sqrt{3} + \sqrt{2})\,^2}{(\,\sqrt{3})\,^2 - (\,\sqrt{2})\,^2}\\\\\frac{(\,\sqrt{3})\,^2 + (\,\sqrt{2})\,^2 + 2(\,\sqrt{3})\,(\,\sqrt{2})\,}{3 - 2}\\\\ \frac{3 + 2 + 2\sqrt{6}}{1}\\\\\ 5 + 2\sqrt{6}

Now, we have

LHS  = RHS

5 + 2\sqrt{6} = a + b\sqrt{6}

On comparing, we get

a = 5

b = 2

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