Question

Find value of a and b​

Answer 1

Given:-

• $\sf{\dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a + b\sqrt{3}}$

To Find:-

• The value of a and b

Solution:-

Let us rationalize the denominator in LHS.

$\sf{\dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}}}$

Multiply the fraction by the conjugate of the denominator

= $\sf{\dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}}\times \dfrac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}}}$

Applying the identity (a + b)(a - b) = a² - b² in the denominator

$\sf{:\implies \dfrac{(5 + 2\sqrt{3})(7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2}}$

$\sf{:\implies \dfrac{5(7 - 4\sqrt{3}) + 2\sqrt{3}(7 - 4\sqrt{3})}{49 - 48}}$

$\sf{:\implies \dfrac{35 - 20\sqrt{3} + 14\sqrt{3} - 24}{1}}$

$\sf{:\implies 11 - 6\sqrt{3}}$

After Rationalizing the denominator we get:-

11 - 6√3 = a + b√3

On Comparing, we get:-

$\sf{a = 11}$

$\sf{b\sqrt{3} = -6\sqrt{3}}$

$\sf{:\implies b = \dfrac{-6\sqrt{3}}{\sqrt{3}}}$

$\sf{:\implies b = \dfrac{-6\not{\sqrt{3}}}{\not{\sqrt{3}}}}$

$\sf{:\implies b = -6}$

$\sf{\therefore The\:value\:of\:a\:is\:11}$

And,

$\sf{The\:value\:of\:b\:is\: -6}$

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