Question

find the value of a and b (question-related to ncert 9th std chapter: 1)

Answer 1

$\huge\sf\pink{Answer}$

☞ Value of a is 0 and b is 1

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$\huge\sf\blue{Given}$

✭ $\sf \dfrac{7+3\sqrt{5}}{3+\sqrt{5}} - \dfrac{7-3\sqrt{5}}{3-\sqrt{5}} = a+b\sqrt{5}$

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$\huge\sf\gray{To \:Find}$

◈ The value of a & b?

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$\huge\sf\purple{Steps}$

Your Answer has been attached!!

So here we use the concept of Rationalising the denominator

Example:-

➝ $\sf \dfrac{a+b}{a-b}$

➝ $\sf \dfrac{1}{a-b} \times \dfrac{a+b}{a+b}$

That is we multiply both the numbers and the denominator with the conjugate of the denominator

Conjugate means the number with the opposite sign here in the example from +ve to -ve

If we have any radical in the denominator what we do is we use this technique so that we can eliminate the radical

Here :-

In the given Question we shall first rationalise the two fractions given,then we shall subtract and equate them to a+b√5 so that we get our final values of a & b

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Answer 2

Question:-

Find a and b if : (7 + 3√5)/(3 + √5) - (7 - 3√5)/(3 - √5) = a + b√5

Answer:-

Given : (7 + 3√5)/(3 + √5) - (7 - 3√5)/(3 - √5) = a + b√5

1st rationalising the bolded part:-

(7 + 3√5)/(3 + √5)

= [(7 + 3√5)(3 - √5)] / [(3 + √5)(3 - √5)]

= [21 - 7√5 + 9√5 - 15] / [3² - (√5)² ]

= (6 + 2√5)/4 ----- (i)

2nd rationalising the underlined part:-

(7 - 3√5)/(3 - √5)

= [(7 - 3√5)(3 + √5)] / [(3 - √5)(3 + √5)]

= [21 + 7√5 - 9√5 - 15] / [3² - (√5)² ]

= (6 - 2√5)/4 -------(ii)

So according to given equation:-

(6 + 2√5)/4 - (6 - 2√5)/4

= ( 6 + 2√5 - 6 + 2√5)/4

= (4√5)/4

= √5

And given:

√5 = a + b√5

=> 0 + 1√5 = a + b√5

So by comparing LHS and RHS, we have

a = 0 and b = 1

Ans. a = 0 and b = 1