Question

Find rational numbers 'a' and 'b' so that -

$\bf \dfrac{5 \: + \: 2 \sqrt{3}}{7 \: + \: 4 \sqrt{3}} = \: a \: + \: b \: \sqrt{3}$
Note !
Irrelevant answers will be deleted on the spot.

Want answer with steps. ​

Answer 1

★ Concept :-

Here the concept of rationalisation has been used. We see that we are given an equation where we need to find the value of two variable terms. Since we aren't given any other equation, so we have relate the terms from the given things. The easiest way to solve such questions is by rationalising and then comparing. Firstly we will rationalise the LHS and then compare it with the RHS.

Let's do it !!

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★ Solution :-

Given,

$\;\bf{\mapsto\;\;\green{\dfrac{5\:+\:2\sqrt{3}}{7\:+\:4\sqrt{3}}\;=\;a\;+\;b\sqrt{3}}}$

From this we get that,

$\;\sf{\rightarrow\;\;\orange{L.H.S.\;=\;\dfrac{5\:+\:2\sqrt{3}}{7\:+\:4\sqrt{3}}}}$

And,

$\;\sf{\rightarrow\;\;\blue{R.H.S.\;=\;a\;+\;b\sqrt{3}}}$

• For rationalisation of L.H.S ::

In order to rationalise L.H.S., we must multiply both numerator and denominator there by (7 - 4√3). This will give,

$\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{5\:+\:2\sqrt{3}}{7\:+\:4\sqrt{3}}\:\times\:\dfrac{7\:-\:4\sqrt{3}}{7\:-\:4\sqrt{3}}}}$

$\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{(5\:+\:2\sqrt{3})(7\:-\:4\sqrt{3})}{(7\:+\:4\sqrt{3})(7\:-\:4\sqrt{3})}}}$

Let's firstly solve the Numerator first.

$\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{35\:-\:20\sqrt{3}\:+\:14\sqrt{3}\:-\:24}{(7\:+\:4\sqrt{3})(7\:-\:4\sqrt{3})}}}$

$\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{11\:-\:6\sqrt{3}}{(7\:+\:4\sqrt{3})(7\:-\:4\sqrt{3})}}}$

Now we know that identity that ::

→ (a + b)(a - b) = a² - b²

• Here a = 7

• Here b = 4√3

By applying this here, we get

$\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{11\:-\:6\sqrt{3}}{(7)^{2}\:-\:(4\sqrt{3})^{2}}}}$

$\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{11\:-\:6\sqrt{3}}{49\:-\:48}}}$

$\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{11\:-\:6\sqrt{3}}{1}}}$

$\;\bf{\Longrightarrow\;\;\red{L.H.S.\;=\;\bf{11\:-\:6\sqrt{3}}}}$

• For value of a and b ::

Now here comes the logic. Let's apply value of L.H.S in the initial equation. So,

$\;\bf{\leadsto\;\;11\:-\:6\sqrt{3}\;=\;a\;+\;b\sqrt{3}}$

Here √3 is constant for second term at both sides.

So this means value of a should be 11 and value of b should be -6.

Then only this condition,

$\;\bf{\leadsto\;\;11\:-\:6\sqrt{3}\;=\;a\;+\;b\sqrt{3}}$

will become true.

This gives us the main answer.

$\;\underline{\boxed{\tt{Value\;\:of\;\:a\;=\;\bf{\purple{11}}}}}$

$\;\underline{\boxed{\tt{Value\;\:of\;\:b\;=\;\bf{\purple{-6}}}}}$

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★ More to know :-

• Rationalisation : It is the process by which the square root term or any complicated term in the denominator of a fraction is reduced by multiplying the numerator and denominator of the fraction by the same term as denominator but opposite in sign.

• L.H.S : Left Hand Side

• R.H.S. : Right Hand Side

Answer 2

Answer:

• Hope it helps you!

Step-by-step explanation:

$\pmb{\underline{\frak{Given:}}}$

$\bf{\dfrac{5+2 \sqrt{3}}{7+4 \sqrt{3}} = \: a+b \sqrt{3}}$

$\pmb{\underline{\sf{According\:to\:the\:question:}}}$

Here, the concept of Rationalisation is used. In this given question, we need to eliminate the √ or radical of the denominator by using some identities. Let's solve this:

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

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

Now, at the denominator side, we are going to use an important algebraic identity:

• $\bf{(a+b)(a-b) = a^2-b^2}$

At the numerator side, we are not using any identity and we are going to just multiply.

$\implies\bf{\dfrac{35-20 \sqrt{3}+14 \sqrt{3}-24}{7^2-(4 \sqrt{3})^2}}$

$\implies\bf{\dfrac{35-6 \sqrt{3}-24}{49- 16 \times 3}}$

$\implies\bf{\dfrac{11-6 \sqrt{3}}{49- 48}}$

$\implies\bf{\dfrac{11-6 \sqrt{3}}{1}}$

$\implies\bf{11-6 \sqrt{3}}$

We can see that our following answer is in the form of $\bf{a+b \sqrt{3}}$ where,

• a = 11

• b = -6

$\pmb{\underline{\frak{Final\:Answer:}}}$

$\boxed{\pink{\frak{a = 11}}}$

$\boxed{\blue{\frak{b = -6}}}$