Question

if 5+2√3 upon 7+4√3 = a-b √3 ​

Answer 1

$\underline{ \underline{ \Large \pmb{\mathit{ {Given:}} }} }$

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

$\underline{ \underline{ \Large \pmb{\mathit{ {To \: calculate:}} }} }$

• Value of a and b.

$\underline{ \underline{ \Large \pmb{\mathit{ {Explication \: of \: steps :}} }} }$

Rationalising the denominator in L.H.S :

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

Rationalising factor of a + b√x is a - b√x, so rationalising factor of 7 + 4√3 is 7 - 4√3. Multiplying it with both denominator and numerator.

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

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

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

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

$\longrightarrow \sf { \dfrac{35 - 20 \sqrt{3} + 14 \sqrt{3} - (8 \times 3)}{49 - (16 \times 3)}}$

$\longrightarrow \sf { \dfrac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 -48}}$

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

$\longrightarrow \sf { \dfrac{11- 20 \sqrt{3} + 14 \sqrt{3} }{1}}$

$\longrightarrow \sf {11+ \sqrt{3} ( - 20 + 14) }$

$\longrightarrow \sf {11 + \sqrt{3} ( - 6) }$

$\longrightarrow \boxed{ \sf {11 -6 \sqrt{3}} }$

R.H.S :

$\longrightarrow \boxed{ \sf {a-b \sqrt{3}} }$

Comparing L.H.S and R.H.S,

$\longrightarrow \boxed{ \sf {a-b \sqrt{3} = 11 -6 \sqrt{3}}}$

We get that,

$\underline{\boxed{\sf{ a = 11}}} \: \red{\bigstar}$

$\underline{\boxed{\sf{ b = 6}}} \: \red{\bigstar}$

So, the value of a is 11 and the value of b is 6.

Points to remember:

• (a + b) and (a - b) are rationalising factors of each other.

• (a + b√x) and (a - b√x) are rationalising factors of each other.

Some Identities:

• (√a)² = a

• √a√b = √ab

• √a/√b = √a/b

• (√a + √b)(√a - √b) = a - b

• (a + √b)(a - √b) = a² - b

• (√a ± √b)² = a ± 2√ab + b

• (√a + √b)(√c + √d) = √ac + √ad + √bc + √bd

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 - b²

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