Question

solve the equation given below

Answer 1

ANSWER:

Given:

• (5 + 2√3)/(7 + 4√3) = a + b√3

To Find:

• Value of a and b.

Solution:

We are given that,

$\implies\dfrac{5+2\sqrt3}{7+4\sqrt3}=a+b\sqrt3$

Now, we will rationalise it.

The rationalising factor of 7 + 4√3 = 7 - 4√3

So,

On multiplying and dividing LHS by 7 - 4√3

$\implies\dfrac{5+2\sqrt3}{7+4\sqrt3}\times\dfrac{7-4\sqrt3}{7-4\sqrt3}=a+b\sqrt3$

Combining fractions,

$\implies\dfrac{(5+2\sqrt3)(7-4\sqrt3)}{(7+4\sqrt3)(7-4\sqrt3)}=a+b\sqrt3$

We know that,

⇒ (x + y)(x - y) = x² - y²

So,

$\implies\dfrac{(5+2\sqrt3)(7-4\sqrt3)}{(7+4\sqrt3)(7-4\sqrt3)}=a+b\sqrt3$

$\implies\dfrac{(5+2\sqrt3)(7-4\sqrt3)}{(7)^2-(4\sqrt3)^2}=a+b\sqrt3$

$\implies\dfrac{(5\times7)+(5\times-4\sqrt3)+(2\sqrt3\times7)+(-2\sqrt3\times4\sqrt3)}{(7)^2-(\sqrt{48})^2}=a+b\sqrt3$

$\implies\dfrac{35-20\sqrt3+14\sqrt3-24}{49-48}=a+b\sqrt3$

$\implies\dfrac{11-6\sqrt3}{1}=a+b\sqrt3$

$\implies 11-6\sqrt3=a+b\sqrt3$

On comaring the terms,

• a = 11
• b = -6

Therefore, the value of a and b is 11 and -6 respectively.

Answer 2

Step-by-step explanation:

$\bf{ \underline{Given-}}$

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

$\bf \underline{To \: find \: out-}$

$\rm{Value \: of \: a \: and \: b \: in \: given \: expression.}$

$\bf \underline{Solution-}$

Given expression

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

The denominator is 7 + 4√3.

We know that

Rationalising factor of a + b√c = a - b√c.

So, the rationalising factor of 7 +4√3 = 7-4√3.

On rationalising the denominator them

$\longrightarrow \: \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

$\longrightarrow \: \frac{(5 + 2 \sqrt{3})(7 - 4 \sqrt{3}) }{(7 + 4 \sqrt{3})(7 - 4 \sqrt{3}) }$

Now, applying algebraic identity in denominator because it is in the form of;

(a+b)(a-b) = a² - b²

Where, we have to put in our expression: a = 7 and b = 4√3 , we get

$\longrightarrow \: \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 {)}^{2} - (4 \sqrt{3} {)}^{2} }$

$\longrightarrow \: \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{49 - 48}$

Subtract 49 from 48 in denominator to get 1.

$\longrightarrow \: \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{1}$

$\longrightarrow \: (5 + 2 \sqrt{3} )(7 - 4 \sqrt{3} )$

Now, multiply both term left side to right side.

$\longrightarrow \: 35 + 14 \sqrt{3} - 20 \sqrt{3} - 8 \sqrt{3 \times 3 }$

$\longrightarrow \: 35 + 14 \sqrt{3} - 20 \sqrt{3} - 24$

$\longrightarrow \: (35 - 24) - 6 \sqrt{3}$

$\longrightarrow \: 11 - 6 \sqrt{3}$

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

On, comparing with R.H.S , we have

a = 11 and b = -6

$\bf \underline{Hence, value \: of \: a = 11 \: and \: b = - 6.}$

Used Formulae:

(a+b)(a-b) = a² - b

Rationalising factor of a + b√c = a - b√c.