Question

square root 3 + 1
------------------------- = a + b *square root 3
square root 3 + 1

find a and b

Answer 1

Step-by-step explanation:

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Answer 2

${\underline{{\text{Question}}}}$

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

${\underline{{\text{Answer}}}}$

a = 2, b = 1

${\underline{{\text{To Find:}}}}$

The value of a and b .

${\underline{{\text{Solution:}}}}$

We have to do nothing just rationalize the denominator and compare the L.H.S and R.H.S in last and we will get the answer.

L.H.S

$\frac{ \sqrt{3} + 1}{ \sqrt{3 - 1} } \\ \\ \frac{( \sqrt{3} + 1)( \sqrt{3} + 1) }{( \sqrt{3} - 1)( \sqrt{3} + 1) } \\ \\ \frac{( \sqrt{3} + 1) {}^{2} }{ ({ \sqrt{3}) }^{2} - {(1)}^{2} } \: \: \: [\text{Using i and ii}] \\ \\ \frac{3 + 1 + 2 .1.\sqrt{3} }{3 - 1} \\ \\ \frac{4 + 2 \sqrt{3} }{2} \\ \\ \frac{2(2 + \sqrt{3}) }{2} \\ \\ 2 + \sqrt{3}$

Comparing the L.H.S and R.H.S I get,

a = 2

b√3 = √3 ⟹ b = 1

Hence, The required value of a is 2 and b is 1.

${\underline{{\text{Important Formula for Algebra}}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy ....... [\text{i}]\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}- (x-y) {}^{2}=4xy\\ \\ {(x + y)}^{3}={x}^{3}+{y}^{3}+ 3xy(x + y) \\ \\(x - y)^{3}={x}^{3}-{y}^{3}- 3xy(x - y)\\ \\(x+y)(x-y) = x^2 - y^2........[\text{ii}]$