Question

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Q Find the values of a and b​

Answer 1

Answer:

a=0, b= 1

Step-by-step explanation:

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

21-7√5+9√5-15 / 9-5 - 21+7√5-9√5-15 / 9-5

6+2√5 / 4 - 6-2√5 / 4

2(3+√5) / 4 - 2(3-√5) / 4

3+√5-3+√5 / 2

√5 = a+√5b

a+ √5b= √5*1 + a*0 (comparing two equations)

hence a=0, b=1

Answer 2

$\large \sf \underline{Solution - }$

$\rm Given \: : \: \dfrac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \dfrac{7 - 3 \sqrt{5} }{3 - \sqrt{5} }= a + \sqrt{5b}$

• Here, we have to find the value of a and b.

➢ Multiplying the conjugate of their denominator to both the fractions.

So,

$\rm \: \implies \dfrac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \dfrac{7 - 3 \sqrt{5} }{3 - \sqrt{5} }= a + \sqrt{5b}$

$\rm \: \implies \dfrac{(7 + 3 \sqrt{5})(3 - \sqrt{5}) }{(3 + \sqrt{5} )(3 - \sqrt{5})} - \dfrac{(7 - 3 \sqrt{5}) (3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5}) }= a + \sqrt{5b}$

We know,

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

Then,

$\rm \: \implies \dfrac{(7 + 3 \sqrt{5})(3 - \sqrt{5}) }{3^{2} - ( \sqrt{5} )^{2} } - \dfrac{(7 - 3 \sqrt{5}) (3 + \sqrt{5})}{3^{2} - ( \sqrt{5} )^{2} } = a + \sqrt{5b}$

$\rm \: \implies \dfrac{(7 + 3 \sqrt{5})(3 - \sqrt{5}) }{9- 5 } - \dfrac{(7 - 3 \sqrt{5}) (3 + \sqrt{5})}{9 - 5 } = a + \sqrt{5b}$

$\rm \: \implies \dfrac{(7 + 3 \sqrt{5})(3 - \sqrt{5}) }{4} - \dfrac{(7 - 3 \sqrt{5}) (3 + \sqrt{5})}{4 } = a + \sqrt{5b}$

Combine both the fractions,

$\rm \: \implies \dfrac{(7 + 3 \sqrt{5})(3 - \sqrt{5}) - (7 - 3 \sqrt{5}) (3 + \sqrt{5})}{4} = a + \sqrt{5b}$

$\rm \: \implies \dfrac{21-7\sqrt{5}+9\sqrt{5}-15-21-7\sqrt{5}+9\sqrt{5}+15}{4}= a + \sqrt{5b}$

$\rm \: \implies \dfrac{4 \sqrt{5} }{4}= a + \sqrt{5b}$

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

$\rm \: : \implies 0 + \sqrt{5 \times 1} = a + \sqrt{5b}$

Therefore,

• The value of a is 0
• The value of b is 1