Question

√3-1÷√3+1 + √3+1÷√3-1= a+√3b find a and b

Answer 1

• What is a rational number ?

  A rational number is of the form $\frac{a}{b}$ where both a and b are integers with a non zero b.

• What is an irrational number ?

  The numbers which can not be expressed in $\frac{a}{b}$ form are called irrational numbers.

• Some other facts :

  All the integers are rational numbers but all the rational numbers aren't necessarily integers.

• What we are going to do?

  While solving the given problem, we need to rationalise the terms so that we get a simplied form consisting of rational and irrational parts. For this, we take LCM of both the denominators and multiply the uncommon term with each of the numerators and then we find product of those terms. After doing mathematical operations (addition and subtraction), we finally find the simplified form. Then we equate the terms from both the sides and after a comparison between rational and irrational terms, we find the value of a and b.

• Solution :

$\underline{\textsf{Let's simplify the given term.}}$

$\textsf{Now,}\:\dfrac{\sqrt{3}-1}{\sqrt{3}+1} + \dfrac{\sqrt{3}+1}{\sqrt{3}-1}$

$\bold{=\frac{(\sqrt{3}-1)(\sqrt{3}-1)+(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}}$

$\bold{=\frac{(3-\sqrt{3}-\sqrt{3}+1)+(3+\sqrt{3}+\sqrt{3}+1)}{3-\sqrt{3}+\sqrt{3}-1}}$

$=\dfrac{4-2\sqrt{3}+4+2\sqrt{3}}{2}$

$=\dfrac{8}{2}$

$= \bold{4}$

$\underline{\textsf{Given that,}}$

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

$\implies \bold{4 = a + \sqrt{3}b}$

$\implies \bold{4.1 + 0.\sqrt{3} = a.1 + b.\sqrt{3}}$

$\underline{\textsf{Comparing both sides, we get}}$

$\boxed{\boxed{\bold{a = 4}}\:\: \& \: \:\boxed{\bold{b = 0}}}$

Answer 2

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