Question

Find the value of a and b in

Answer 1

Answer:

a =7

b=5

PLEASE MARK AS BRAINLIEST AND A THANKS PLEASE

Answer 2

Given :

• $\tt\:\frac{2-\sqrt{5}}{2+\sqrt{5}}-\frac{2+\sqrt{5}}{2-\sqrt{5}}=a+b\sqrt{5}$

‎

‎

To find :

• The value of a and b

‎

‎

Concept :

In order to calculate the value of a and b we will first rationalize both the denominators of the ratios then comparing LHS with RHS we will get the required value of a and b.

‎Rationalization is to be done by multiplying the ratio by the conjugate of the denominator and by using some identity.

‎

Identities to be used :

❒ $\sf\color{purple}{(a+b) ^{2}=a^{2}+2ab+b^{2}}$

❒ $\sf\color{purple}{(a-b) ^{2}=a^{2}-2ab+b^{2}}$

❒ $\sf\color{purple}{(a+b)(a-b)=(a)^{2}-(b)^{2}}$

Hope am clear let's solve :D~‎

‎

‎

Solution :

According to the question ,

$\tt\:\frac{2-\sqrt{5}}{2+\sqrt{5}}-\frac{2+\sqrt{5}}{2-\sqrt{5}}=a+b\sqrt{5}$

Rationalising the denominators of both the ratio

• Conjugate of 2 + √5 = $\bf\:2-\sqrt{5}$

• Conjugate of 2 - √5 = $\bf\:2+\sqrt{5}$

$\tt\implies\:\frac{2-\sqrt{5}}{2+\sqrt{5}} \times \frac{2-\sqrt{5}}{2-\sqrt{5}} -\frac{2+\sqrt{5}}{2-\sqrt{5}}\times \frac{2+\sqrt{5}}{2+\sqrt{5}} =a+b\sqrt{5}$

Multiplying the ratios in LHS

$\tt\implies\:\frac{(2-\sqrt{5})^{2}}{(2)^{2}-(\sqrt{5})^{2}} -\frac{(2+\sqrt{5})^{2}}{(2)^{2}-(\sqrt{5})^{2}}=a+b\sqrt{5}$

Using the identities as mentioned earlier

$\tt\implies\:\frac{(2) ^{2}-2 \times 2 \times \sqrt{5} +(\sqrt{5})^{2}}{(2) ^{2}-(\sqrt{5})^{2}}-\frac{(2) ^{2}+2 \times 2 \times \sqrt{5} +(\sqrt{5})^{2}}{(2) ^{2}-(\sqrt{5})^{2}}=a+b\sqrt{5}$

$\tt\implies\:\frac{4-4\sqrt{5}+5}{4-5}-\frac{4+4\sqrt{5}+5}{4-5}=a+b\sqrt{5}$

$\tt\implies\:\frac{4-4\sqrt{5}+5}{-1}-\frac{4+4\sqrt{5}+5}{-1}=a+b\sqrt{5}$

Taking LCM

$\tt\implies\:\frac{(4-4\sqrt{5}+5)-(4+4\sqrt{5}+5)}{-1}=a+b\sqrt{5}$

Multiplying the signs and removing the brackets

$\tt\implies\:\frac{4-4\sqrt{5}+5-4-4\sqrt{5}-5}{-1}=a+b\sqrt{5}$

Arranging the numbers

$\tt\implies\:\frac{4+5-4-5-4\sqrt{5}-4\sqrt{5}}{-1}=a+b\sqrt{5}$

Proceeding with simple calculation

$\tt\implies\:\frac{9-9-8\sqrt{5}}{-1}=a+b\sqrt{5}$

$\tt\implies\:\frac{\cancel{9}-\cancel{9}-8\sqrt{5}}{-1}=a+b\sqrt{5}$

$\tt\implies\:\frac{-8\sqrt{5}}{-1}=a+b\sqrt{5}$

$\tt\implies\:\frac{\cancel{-}8\sqrt{5}}{\cancel{-}1}=a+b\sqrt{5}$

$\tt\implies\:\frac{8\sqrt{5}}{1}=a+b\sqrt{5}$

$\tt\implies\:8\sqrt{5}=a+b\sqrt{5}$

Comparing LHS and RHS we get :

• a = $\large{\mathfrak\red{0}}$

• b = $\large{\mathfrak\red{8}}$

______________________

‎