Question

please solve it and ​

Answer 1

Step-by-step explanation:

The value of (a+b)a is 20

The given equation is

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

5

−

3

5

+

3

=a+b

15

\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=a+b\sqrt{15}

5

−

3

5

+

3

×

5

+

3

5

+

3

=a+b

15

\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5})^2-(\sqrt{3})^2}=a+b\sqrt{15}

(

5

)

2

−(

3

)

2

(

5

+

3

)

2

=a+b

15

\frac{5+2\sqrt{15}+3}{5-3}=a+b\sqrt{15}

5−3

5+2

15

+3

=a+b

15

\frac{8+2\sqrt{15}}{2}=a+b\sqrt{15}

2

8+2

15

=a+b

15

4+\sqrt{15}=a+b\sqrt{15}4+

15

=a+b

15

On comparing both sides.

a=4a=4

b=1b=1

The value of (a+b)a is

(a+b)a=(4+1)4=5\times 4=20(a+b)a=(4+1)4=5×4=20

Therefore the value of (a+b)a is 20.

Answer 2

       $\underline{\bold{Answer:}}$

       $a= 4 \text{ and }b = 1$

       $\underline{\bold{Step-by-step-explaination:}}$

       $\text{The given expression is: }\dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3} = a + b\sqrt{15}$

       $\text{Taking L.H.S}$

$\implies \dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3}$

       $\text{Multiplying and dividing by }(\sqrt5+\sqrt3)$

 $\implies \dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3}\text{ x }\dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}$

       $\text{We know that, }(a+b)(a-b) = a^{2} - b^{2}$      

$\implies \dfrac{(\sqrt5 + \sqrt3)^{2} }{(\sqrt5)^{2} - (\sqrt3)^{2} }$

       $\text{Simplifying...}$

$\implies \dfrac{(\sqrt5 + \sqrt3)^{2} }{5 - 3}$

       $\text{We know that, }(a+b)^{2} = a^{2} + b^{2} + 2ab$

$\implies \dfrac{(\sqrt5)^{2} + (\sqrt3)^{2} + 2(\sqrt5)(\sqrt3) }{5 - 3}$

       $\text{Simplifying...}$

$\implies \dfrac{5 + 3 + 2\sqrt{15}}{2}$

       $\text{Simplifying...}$

$\implies \dfrac{8 + 2\sqrt{15}}{2}$

       $\text{Taking 2 common in numerator and simplifying...}$

$\implies 4+\sqrt{15}$

       $\text{Now comparing with R.H.S}$

$\implies 4+\sqrt{15} = a + b\sqrt{15}$

  $\therefore \text{ }\text{ }\boxed{a= 4 \text{ and }b = 1}$