Question

.if A and B are rational numbers and
$2 \sqrt{5} + \sqrt{3} \div 2 \sqrt{5} - \sqrt{3} + 2 \sqrt{5} - 3 \div 2 \sqrt{5} + 3 = a + \sqrt{15} b$

find the values of a and b ​

Answer 1

Correct Question :-

[(2√5 + √3) / (2√5 - √3)] + [(2√5 - √3) / (2√5 + √3)] = a + √15b

To Find :-

• value of a & b ?

Solution :-

Lets First Rationalize The first Part :-

→ [(2√5 + √3) / (2√5 - √3)]

Multiply & Divide by (2√5 + √3) we get,

→ [(2√5 + √3) / (2√5 - √3)] * [ (2√5 + √3) / (2√5 + √3) ]

Numerator Become (a + b)(a+b) = (a + b)² & Denominator Becomes (a + b)(a - b) = a² - b² .

So,

→ [ (2√5 + √3)² ] / [ (2√5)² - (√3)² ]

Using (a + b)² = a² + b² + 2ab in Numerator,

→ [ (20 + 3 + 4√15) / (20 - 3) ]

→ (23 + 4√15) / 17 ----------- Equation (1)

___________________

Rationalize Second Part Now,

→ [(2√5 - √3) / (2√5 + √3)]

Multiply & Divide by (2√5 - √3)

→ [ (2√5 - √3)² / {(2√5)² - (√3)²} ]

→ [ (20 + 3 - 4√15) / 17 ]

→ (23 - 4√15) / 17 ----------- Equation (2)

___________________

Adding Equation (1) & (2) Now, we get,

→ [(23 + 4√15) / 17] + [(23 - 4√15) / 17] = a + √15b

→ [ (23 + 23 + 4√15 - 4√15) / 17 ] = a + √15b

→ (46/17) = a + √15b

Comparing we get,

→ a = (46/17) .

→ b = 0 .

Answer 2

The question is somehow wrong. The correct question is:-

[(2√5 + √3) / (2√5 - √3)] + [(2√5 - √3) / (2√5 + √3)] = a + √15b

Given:-

• 'a' and 'b' are rational numbers.

• [(2√5 + √3) / (2√5 - √3)] + [(2√5 - √3) / (2√5 + √3)] = a + √15b

To find:-

The values of 'a' and 'b'.

Solution:-

$\frac{(2\sqrt{5} + \sqrt{3} ) }{(2\sqrt{5} - \sqrt{3} )} +\frac{ (2\sqrt{5} - \sqrt{3} ) }{(2\sqrt{5} + \sqrt{3} )} = a + \sqrt{15} b$

$\frac{(2\sqrt{5} + \sqrt{3} )(2\sqrt{5} + \sqrt{3} ) }{(2\sqrt{5} - \sqrt{3})(2\sqrt{5} + \sqrt{3} )}+\frac{(2\sqrt{5} - \sqrt{3} )(2\sqrt{5} - \sqrt{3} )}{(2\sqrt{5} + \sqrt{3} )(2\sqrt{5} - \sqrt{3} )} = a + \sqrt{15} b$

$\frac{(2\sqrt{5} + \sqrt{3} )^2}{2\sqrt{5}^2-\sqrt{3}^2}+\frac{(2\sqrt{5} - \sqrt{3} )^2}{2\sqrt{5}^2-\sqrt{3}^2 }=a+\sqrt{15}b$

$\frac{(2\sqrt{5})^2+\sqrt{3}^2+2*\sqrt{3}*2\sqrt{5} }{20-3}+\frac{(2\sqrt{5})^2 - 2*2\sqrt{5}*\sqrt{3}+(\sqrt{3})^2}{20-3}=a+\sqrt{15}b$

$\frac{20+3+4\sqrt{15} }{17}+\frac{20-4\sqrt{15}+3}{17}=a+\sqrt{15}b$

$\frac{23+4\sqrt{15} }{17}+\frac{23-4\sqrt{15}}{17}=a+\sqrt{15}b$

$\frac{23+4\sqrt{15}+23-4\sqrt{15}}{17} = a+15\sqrt{b}$

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

here, 46/17 can be written as 46/17 + 0.

Comparing LHS and RHS, we get:

$\boxed{a = \frac{46}{17} }$

15√b = 0

⇒√b = 0

⇒$\boxed{b = 0}$