Question

If a + b√3 =√3-1/√3+1, find the value of a and b​

Answer 1

EXPLANATION.

To find Value of a and b.

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

$\sf : \implies \: \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1} = a \: + \: b \sqrt{3} \\ \\ \sf : \implies \: rationalise \: the \: term \\ \\ \sf : \implies \: \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} \times \frac{ \sqrt{3 } - 1 }{ \sqrt{3} - 1} \\ \\ \sf : \implies \: \frac{ (\sqrt{3} - 1) {}^{2} }{( \sqrt{3} ) {}^{2} -( 1) {}^{2} }$

$\sf : \implies \: \dfrac{(3 + 1 - 2 \sqrt{3} )}{2} \\ \\ \sf : \implies \: \frac{4 - 2 \sqrt{3} }{2} \\ \\ \sf : \implies \: \frac{2(2 - \sqrt{3} )}{2} \\ \\ \sf : \implies \: 2 - \sqrt{3} = a \: + \: b \: \sqrt{3}$

$\sf : \implies \: \green{{ \underline{value \: of \: a \: = 2 \: \: and \: \: \: b \: = - 1}}}$

Answer 2

Given:-

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

To find:-

The value of a and b

Solution:-

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

First we will rationalise the denominator

Here, denominator is √3 + 1 , so it's conjugate will be √3 - 1

Multiplying √3 - 1 in numerator and denominator will imply,

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

$\rm\dashrightarrow a+b\sqrt{3}=\dfrac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}$

Since, $\rm (a-b)^2=a^2+b^2-2ab \ and \ (a+b)(a-b)=a^2-b^2$, Therefore,

$\rm\dashrightarrow a+b\sqrt{3}=\dfrac{(\sqrt{3})^2+(1)^2-2\times\sqrt{3}\times 1}{(\sqrt{3})^2-(1)^2}$

$\rm\dashrightarrow a+b\sqrt{3}=\dfrac{3+1-2\sqrt{3}}{3-1}$

$\rm\dashrightarrow a+b\sqrt{3}=\dfrac{4-2\sqrt{3}}{2}$

$\rm\dashrightarrow a+b\sqrt{3}=\dfrac{2(2-1\sqrt{3})}{2}$

$\rm\dashrightarrow a+b\sqrt{3}=2-1\sqrt{3}$

$\rm\dashrightarrow a+b\sqrt{3}=2+(-1\sqrt{3})$

On comparing, we get

a = 2 and b = -1