Question

if a and b are rational number, find a and b
√5-2/√5+2 - √5+2/√5-2 = a + b√5​

Answer 1

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Answer 2

Answer :

• a = 0
• b = -8

___________________

As per the provided information in the given question, we have to find the value of a and b. Here,

$\longmapsto \rm {\dfrac{\sqrt{5}-2}{\sqrt{5}+2} - \dfrac{\sqrt{5}+2}{\sqrt{5}-2} = a + b\sqrt{5}}$

Solving L.H.S :

Here, we have two expressions in L.H.S:

$\longmapsto \bf { Exp. \; 1 : \dfrac{\sqrt{5}-2}{\sqrt{5}+2} }$

$\longmapsto \bf { Exp. \; 2 : \dfrac{\sqrt{5}+2}{\sqrt{5}-2} }$

Rationalising the denominator of both expressions.

In order to rationalise the denominator of any fraction, we need to multiply the rationalising fact of the denominator with both the numerator and the denominator of the fraction.

Rationalising the denominator of first expression :

$\longmapsto \bf { \dfrac{\sqrt{5}-2}{\sqrt{5}+2} }$

Here, denominator is in the form of (a + b), rationalising factor of (a + b) is (a - b). So, the rationalising factor of (√5 + 2) is (√5 - 2). Multiplying (√5 - 2) with both the numerator and the denominator of the fraction.

$\longmapsto \rm { \dfrac{\sqrt{5}-2}{\sqrt{5}+2} \times \dfrac{\sqrt{5}-2}{\sqrt{5}-2} }$

Rearranging the terms.

$\longmapsto \rm { \dfrac{(\sqrt{5}-2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)} }$

Now by using the two identities given below,

$\bf {(a-b)^2 = a^2 + b^2 - 2ab}$

$\bf {(a+b)(a-b) = a^2 - b^2}$

$\longmapsto \rm { \dfrac{(\sqrt{5})^2 +(2)^2-2(2\sqrt{5}}{(\sqrt{5})^2-(2)^2} }$

Simplifying further.

$\longmapsto \rm { \dfrac{5 +4-4\sqrt{5}}{5-4} }$

Performing addition in numerator and subtraction in denominator.

$\longmapsto \rm { \dfrac{9-4\sqrt{5}}{1} }$

Now, we can write it as,

$\longmapsto \underline { \bf { 9-4\sqrt{5} } }$

Rationalising the denominator of 2nd expression :

$\longmapsto \bf { \dfrac{\sqrt{5}+2}{\sqrt{5}-2} }$

Here, denominator is in the form of (a - b), rationalising factor of (a - b) is (a + b). So, the rationalising factor of (√5 - 2) is (√5 + 2). Multiplying (√5 + 2) with both the numerator and the denominator of the fraction.

$\longmapsto \rm { \dfrac{\sqrt{5}+2}{\sqrt{5}-2} \times \dfrac{\sqrt{5}+2}{\sqrt{5}+2} }$

Rearranging the terms.

$\longmapsto \rm { \dfrac{(\sqrt{5}+2)^2}{(\sqrt{5}-2)(\sqrt{5}+2)} }$

Now by using the two identities given below,

$\bf {(a+b)^2 = a^2 + b^2 + 2ab}$

$\bf {(a+b)(a-b) = a^2 - b^2}$

$\longmapsto \rm { \dfrac{(\sqrt{5})^2 +(2)^2+2(2\sqrt{5}}{(\sqrt{5})^2-(2)^2} }$

Simplifying further.

$\longmapsto \rm { \dfrac{5 +4+4\sqrt{5}}{5-4} }$

Performing addition in numerator and subtraction in denominator.

$\longmapsto \rm { \dfrac{9+4\sqrt{5}}{1} }$

Now, we can write it as,

$\longmapsto\underline{ \bf { 9+4\sqrt{5}} }$

$\rule{200}2$

No, we have :

$\longmapsto \bf {\dfrac{\sqrt{5}-2}{\sqrt{5}+2} - \dfrac{\sqrt{5}+2}{\sqrt{5}-2} = a + b\sqrt{5}}$

$\longmapsto \rm { (9-4\sqrt{5} )-(9+4\sqrt{5})= a + b\sqrt{5}}$

Removing the brackets,

$\longmapsto \rm { 9-4\sqrt{5} -9-4\sqrt{5}= a + b\sqrt{5}}$

$\longmapsto \rm { -4\sqrt{5} -4\sqrt{5}= a + b\sqrt{5}}$

Taking √5 as common,

$\longmapsto \rm {( -4-4)\sqrt{5} = a + b\sqrt{5}}$

$\longmapsto \rm {-8\sqrt{5} = a + b\sqrt{5}}$

Comparing L.H.S and R.H.S, we get :

$\longmapsto \bf {\red{0} + (\red{-8})\sqrt{5} = \red{a + b}\sqrt{5}}$

∴ Value of a is 0 and value of b is -8.