Question

find a and b if 2√5+√3÷2√5-√3+2√5-√3÷2√5+√3=a+b√15

Answer 1

Answer :-

$\boxed{\bf{a = \dfrac{46}{17} \: and \: b = 0}}$

Explanation :-

Given :- $\sf{ \dfrac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - \sqrt{3}} + \dfrac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} + \sqrt{3} } = a + b \sqrt{15}}$

To find :- values of a, b

Solution :-

$\sf{ \dfrac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - \sqrt{3}} + \dfrac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} + \sqrt{3} } = a + b \sqrt{15}}$

Let $\sf{x= \dfrac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} -\sqrt{3}} }$

Let $\sf{y= \dfrac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} + \sqrt{3}} }$

Consider x

Rationalize the denomintors

The rationalizing factor of 2√5 - √3 is 2√5 + 3√2. So, multiply both numerator and denominator.

$\dfrac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - \sqrt{3} } \times \dfrac{2\sqrt{5} + \sqrt{3} }{2 \sqrt{5} + \sqrt{3} }$

$= \dfrac{ {(2 \sqrt{5} + \sqrt{3}) }^{2} }{ {(2 \sqrt{5}) }^{2} - {( \sqrt{3}) }^{2} }$

[Since x² + y² = (x + y)(x - y) ]

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

[Since (x + y)² = x² + y² + 2xy]

$= \dfrac{4(5) + 3 + 4 \sqrt{15} }{20 - 3}$

$= \dfrac{20 + 3 + 4 \sqrt{15} }{17}$

$= \dfrac{23 + 4 \sqrt{15} }{17}$

Consider y

The rationalizing factor of 2√5 + √3 is 2√5 - 3√2. So, multiply both numerator and denominator.

$\dfrac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} + \sqrt{3} } \times \dfrac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} - \sqrt{3} }$

$= \dfrac{ {(2 \sqrt{5} - \sqrt{3})}^{2} }{ {( 2\sqrt{5})}^{2} - {( \sqrt{3}) }^{2} }$

[Since x² + y² = (x + y)(x - y) ]

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

[Since (x - y)² = x² + y² - 2xy]

$= \dfrac{4(5) + 3 - 4 \sqrt{15} }{20-3}$

$= \dfrac{20 + 3 - 4 \sqrt{15} }{17}$

$= \dfrac{23 - 4 \sqrt{15} }{17}$

Now find the value of (x + y)

$\dfrac{23 + 4 \sqrt{15} }{17} + \dfrac{23 - 4 \sqrt{15} }{17}$

$= \dfrac{46}{17}$

Equating a corresponding irrational and rational numbers we have,

$\dfrac{46}{17} = a + b \sqrt{15}$

$a = \dfrac{46}{17}$

$b \sqrt{15} = 0$

$b = 0 \sqrt{15}$

$b = 0$

$\boxed{\bf{a = \dfrac{46}{17} \: and \: b = 0}}$