Question

√3-1/√3+1 =a+b√3

solve it​

Answer 1

Answer:

$a=2, b=-1$

Step-by-step explanation:

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

$LHS=RHS$ $($∵ Given$)$

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

On rationalising the denominator,

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

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

⇒ $\frac{(\sqrt{3}- 1)^2}{(\sqrt{3})^2 -(1)^2}$  $($∵ $(a+b)(a-b)=a^2-b^2)$

⇒ $\frac{(\sqrt{3}- 1)^2}{3-1}$

⇒ $\frac{(\sqrt{3}- 1)^2}{2}$

⇒ $\frac{(\sqrt{3})^2+ (1)^2-(2)(\sqrt{3})(1) }{2}$ $($∵ $(a-b)^2=a^2+b^2-2ab)$

⇒ $\frac{3+ 1-2\sqrt{3} }{2}$

⇒ $\frac{4-2\sqrt{3} }{2}$

⇒ $\frac{2(2-\sqrt{3}) }{2}$

⇒ $2-\sqrt{3}$

$2-\sqrt{3}=a+b\sqrt{3}$ $($∵ $LHS=RHS)$

∴ The values of $a$ and $b$ are $2$ and $-1$ respectively.