Question

If root 3 minus 1 upon root 3 + 1 + root 3 + 1 upon root 3 - 1 = A + root 3 B find the value of a and b

Answer 1

Answer:

a = 4 and b = 0

Step-by-step explanation:

$\frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} + \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} = a + \sqrt{3} b$

Rationalizing the denominators we get:

L.H.S,

$= \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1 }{ \sqrt{3} - 1} + \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 }$

$= \frac{ {( \sqrt{3}) }^{2} + {(1)}^{2} - 2. \sqrt{3} .1 }{ {( \sqrt{3}) }^{2} - {(1)}^{2} } + \frac{ {( \sqrt{3} )}^{2} + {(1)}^{2} + 2. \sqrt{3} .1}{ {( \sqrt{3} )}^{2} - {(1)}^{2} }$

$= \frac{3 + 1 - 2 \sqrt{3} }{3 - 1} + \frac{3 + 1 + 2 \sqrt{3} }{3 - 1} \\ \\ = \frac{4 - 2 \sqrt{3} }{2} + \frac{4 + 2 \sqrt{3} }{2} \\ \\ = 2 - \sqrt{3} + 2 + \sqrt{3} \\ \\ = 2 + 2 \\ \\ \bf = 4$

Comparing L.H.S and R.H.S we get:

4 = a + b√3

a = 4 and b = 0

Answer 2

Hey there

refer to attachment