Question

I have a doubt in this... Also step by step explanation please

Answer 1

a = 2 and b = -1

Step-by-step explanation:

en: \frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}3+13−1=a+b3

To find: vale of a & b

We find value of a & b by rationalizing the denominator of LHS and then equating with RHS

Consider,

LHS

=\frac{\sqrt{3}-1}{\sqrt{3}+1}=3+13−1

=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=3+13−1×3−13−1

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

=\frac{(\sqrt{3})^2+(1)^2-2\sqrt{3}}{(\sqrt{3})^2-(1)^2}=(3)2−(1)2(3)2+(1)2−23

=\frac{3+1-2\sqrt{3}}{3-1}=3−13+1−23

=\frac{4}{2}-\frac{2\sqrt{3}}{2}=24−223

=2-\sqrt{3}=2−3

Now equating with RHS = a + b√3

we get a = 2 7  b = -1

Therefore, a = 2 and b = -1