Question

If both a and b are rational numbers, find the values of a and b of the following equality
√5+√3/√5-√3=a+b√15
​

Answer 1

Given :

$\star \: \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a \: + \: b \sqrt{15}$

To find :

• The value of a and b .

Solution :

$\implies \: \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

Rationalise the denominator :

$\implies \: \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \\ \\ \implies \: \frac{ {( \sqrt{5} + \sqrt{3}) }^{2} }{ {( \sqrt{5} })^{2} - ( { \sqrt{3} }) ^{2} } \\ \\ \implies \: \frac{ { \sqrt{5} }^{2} + { \sqrt{3} }^{2} + 2 \sqrt{5} \sqrt{3} }{5 - 3} \\ \\ \implies \: \frac{5 + 3 + 2 \sqrt{15} }{2} \\ \\ \implies \: \frac{8 + 2 \sqrt{15} }{2} \\ \\ \implies \: \frac{ \cancel{2}(4 + \sqrt{15}) }{ \cancel{2}} \\ \\ \implies \: 4 + \sqrt{15}$

Now , compare with a+b√15 we get ,

• a = 4 and b = 1 .

Formula :

$\star \boxed{ \red{\bold{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}} \\ \\ \star \boxed{ \bold { \blue {(x + y)(x - y) = {x}^{2} - {y}^{2} }}}$

Answer 2

Step-by-step explanation:

Given :

\begin{lgathered}\star \: \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a \: + \: b \sqrt{15} \\\end{lgathered}

⋆

5

−

3

5

+

3

=a+b

15

To find :

• The value of a and b .

Solution :

\begin{lgathered}\implies \: \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \\\end{lgathered}

⟹

5

−

3

5

+

3

Rationalise the denominator :

\begin{lgathered}\implies \: \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \\ \\ \implies \: \frac{ {( \sqrt{5} + \sqrt{3}) }^{2} }{ {( \sqrt{5} })^{2} - ( { \sqrt{3} }) ^{2} } \\ \\ \implies \: \frac{ { \sqrt{5} }^{2} + { \sqrt{3} }^{2} + 2 \sqrt{5} \sqrt{3} }{5 - 3} \\ \\ \implies \: \frac{5 + 3 + 2 \sqrt{15} }{2} \\ \\ \implies \: \frac{8 + 2 \sqrt{15} }{2} \\ \\ \implies \: \frac{ \cancel{2}(4 + \sqrt{15}) }{ \cancel{2}} \\ \\ \implies \: 4 + \sqrt{15}\end{lgathered}

⟹

5

−

3

5

+

3

×

5

+

3

5

+

3

⟹

(

5

)

2

−(

3

)

2

(

5

+

3

)

2

⟹

5−3

5

2

+

3

2

+2

5

3

⟹

2

5+3+2

15

⟹

2

8+2

15

⟹

2

2

(4+

15

)

⟹4+

15

Now , compare with a+b√15 we get ,

• a = 4 and b = 1 .

Formula :

\begin{lgathered}\star \boxed{ \red{\bold{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}} \\ \\ \star \boxed{ \bold { \blue {(x + y)(x - y) = {x}^{2} - {y}^{2} }}}\end{lgathered}

⋆

(x+y)

2

=x

2

+y

2

+2xy

⋆

(x+y)(x−y)=x

2

−y

2