Question

Question :-

If a and b are rational numbers and ( 4 + 3√5)/(4-3√5) = a + b, find the values of a and b

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Answer 1

Solution :-

We have

$\frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = \frac{(4 + 3 \sqrt{5}) }{(4 - 3 \sqrt{5}) } \times \frac{(4 + 3 \sqrt{5}) }{(4 + 3 \sqrt{5} )} \\ \\ = \frac{(4 + 3 \sqrt{5})^{2} }{ ({4})^{2} - (3 \sqrt{5})^{2} } = \frac{(4 + 3 \sqrt{5} )^{2} }{16 - 45} = \frac{(4 + 3 \sqrt{5}) ^{2} }{ - 29} \\ \\ = \frac{(4)^{2} + (3 \sqrt{5})^{2} + 2 \times 4 \times 3 \sqrt{5} }{ - 29} \\ \\ = \frac{(16 + 45 + 24 \sqrt{5}) }{ - 29} \\ \\ = \frac{(61 + 24 \sqrt{5}) }{ - 29} \\ \\ = ( \frac{ - 61}{29}) + ( \frac{ - 24}{29} ) \sqrt{5} . \\ \\ \therefore \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = a + b \sqrt{5} \leftrightarrow ( \frac{ - 61}{29} ) + (\frac{ - 24}{29} ) \sqrt{5} = a + b \sqrt{5} . \\ \\ \therefore a = - \frac{61}{29} \: and \: \ b = - \frac{24}{29} .$

Answer 2

Correct Question

If a and b are rational numbers and $\dfrac{4 + 3\sqrt{5} }{4-3\sqrt{5} }$ = a + b√5, find the values of a and b.

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Answer

We need to first rationalize the denominator.

So first we will multiply the numerator and denominator with the conjugate of the denominator. So while finding the conjugate we just need to change the middle sign of the expression.

Conjugate of denominator → 4 + 3√5

$\sf \implies \dfrac{4 + 3\sqrt{5} }{4-3\sqrt{5} } \times \dfrac{4+3\sqrt{5}}{4+3 \sqrt{5}}$

$\sf \implies \dfrac{(4 + 3\sqrt{5} )^{2}}{(4-3\sqrt{5} )(4+3 \sqrt{5})}$

Now we need to use algebraic identities to simplify. Here we will use two of them -:

• (a + b)² = a² + 2ab + b²
• (a + b)(a - b) = a² - b²

$\sf \implies \dfrac{(4 + 3\sqrt{5} )^{2}}{(4-3\sqrt{5} )(4+3 \sqrt{5})}$

$\sf \implies \dfrac{(4)^{2}+2(4)(3\sqrt{5})+ (3\sqrt{5})^{2}}{(4)^{2}-(3\sqrt{5} )^{2}}$

$\sf \implies \dfrac{16+24\sqrt{5}+ 45}{16-45}$

$\sf \implies \dfrac{61+24\sqrt{5}}{16-45}$

$\sf \implies \dfrac{61+24\sqrt{5}}{-29}$

Now we will make the denominator positive by multiplying -1 to the numerator and denominator.

$\sf \implies \dfrac{61+24\sqrt{5}}{-29}\times \dfrac{-1}{-1}$

$\sf \implies \dfrac{-61-24\sqrt{5}}{29}$

Now we have rationalized the given real number. So let's find the value of a and b respectively.

∴ a = $\bf \dfrac{-61}{29}$ and b = $\bf \dfrac{-24}{29}$

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