Question

If a and b are rational no's and 2+√3÷2-√3=a+b√3 find the value of a square and bsquare

Answer 1

Hey there !

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Given :

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

To find the value of a and b, we need to simply LHS by rationalising it's denominator.

we have ;

$\bold{ \frac{2 + \sqrt{3} }{2 - \sqrt{3} } = a + b \sqrt{3} } \\ \\ \\ \bold{\frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times\frac{2 + \sqrt{3} }{2 + \sqrt{3} }} \\ \\ \\ \bold{ \frac{(2 + \sqrt{3}) {}^{2} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } } \\ \\ \\ \bold{ \frac{4 + 4 \sqrt{3} + 3}{4 - 3} } \\ \\ \\ \bold{ \frac{7 + 4 \sqrt{3} }{1} } \\ \\ \\ \bold{ 7 + 4 \sqrt{3} }$

Now, on comparing LHS with RHS, we get ;

$\bold{7 + 4 \sqrt{3} = a + b \sqrt{3} }$

Therefore,

$\bold{a = 7 \: \: and \: \: b = 4}$

Now, coming to the question :

$\bold{a {}^{2} + b{}^{2}}$

=> (7)² + (4)²

=> 49 + 16

=> 65

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Thanks for the question !

Answer 2

$\mathfrak{ \underline{ \underline{ \huge{ \star \: heya \: buddy \star}}}}$

Given:

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

To solve this question, we need to rationalise LHS first. So, here we go!

$\frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ = \frac{ {(2 + \sqrt{3}) }^{2} }{4 - 3} \\ = \frac{4 + 4 \sqrt{3} + 3}{1} \\ = 7 + 4 \sqrt{3}$

Now, comparing the LHS and RHS,

$7 + 4 \sqrt{3} = a + b \sqrt{3}$

As visible now,

$a = 7$

$b = 4$

Thus,

${a} ^{2} + {b} ^{2} \\ = {7} ^{2} +{4} ^{2} \\ = 49 + 16 \\ = 65$

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