Question

What are the value of a and b respectively, if (3+3√5)/(3+√5) - (7-3√5)/(3-√5) = a+√5b ?​

Answer 1

Answer:

$\displaystyle a = 0 \text{ and } b = 1$

Step-by-step explanation:

We are given that:

$\displaystyle \frac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3\sqrt{5}}{ 3 - \sqrt{5}} = a + \sqrt{5b}$

And we want to determine the values of a and b.

Simplify the fractions. We can multiply each by the conjugate of its denominator:

$\displaystyle \frac{7 + 3\sqrt{5}}{3 + \sqrt{5}} \left(\frac{3-\sqrt{5}}{3-\sqrt{5}}\right) - \frac{7 - 3\sqrt{5}}{3 - \sqrt{5}} \left( \frac{ 3 + \sqrt{5}}{3 + \sqrt{5}}\right)$

Simplify:

$\displaystyle = \frac{(7 + 3\sqrt{5})(3 - \sqrt{5})}{(3)^2 - (\sqrt{5})^2} - \frac{(7 - 3\sqrt{5})(3 + \sqrt{5})}{(3)^2 - (\sqrt{5})^2} \\ \\ \\ = \frac{(21 - 7\sqrt{5} + 9\sqrt{5} - 3(5))}{(9) - (5)} - \frac{(21+ 7\sqrt{5} - 9\sqrt{5} -3(5))}{(9) - (5)} \\ \\ \\ = \frac{21 + 2\sqrt{5} - 15}{4} - \frac{21 -2\sqrt{5}-15}{4} \\ \\ \\ = \frac{(6 + 2\sqrt{5})-(6-2\sqrt{5})}{4} \\ \\ \\ = \frac{4\sqrt{5}}{4} \\ \\ = \sqrt{5} = 0 + \sqrt{5(1)}$

In conclusion, a = 0 and 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, we notice that

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