Question

1+√48 upon 5 √3+4√2-√72-√108+√8+2=a+b√3

Answer 1

Answer:

14 + 9 √3

Step-by-step explanation:

Given 1+√48 upon 5 √3+4√2-√72-√108+√8+2=a+b√3

So 1 + √48 /  5 √3+4√2-√72-√108+√8+2

 1 + 4√3 / 5√3 + 4√2 - 6√2 - 6√3 + 2√2 + 2

 1 + 4√3 / -√3 + 2

 1 + 4√3 / 2 - √3

Now rationalizing the denominator we get

1 + 4√3 / 2 - √3 x 2 + √3 / 2 + √3

(1 + 4√3)(2 + √3) / 4 - 3 (because (a + b)(a - b) = a^2 - b^2

2 + 8√3 + √3 + 12 / 1

14 + 9√3

So now it is in the form of a + b√3

Answer 2

GIVEN :-

$\rm{\dfrac{1+\sqrt{48}}{5\sqrt{3} +4\sqrt{2}-\sqrt{72}-\sqrt{108} +\sqrt{8} +2} = a+b\sqrt{3}}$

To Find:-

The Value of a and b.

Now,

First we will will simplify it and then we will Rationalise it

$\implies\rm{\dfrac{1+\sqrt{48}}{5\sqrt{3} +4\sqrt{2}-\sqrt{72}-\sqrt{108} +\sqrt{8} +2} = a+b\sqrt{3}}$

$\implies\rm{\dfrac{1+4\sqrt{3}}{5\sqrt{3} +4\sqrt{2}-6\sqrt{2}-6\sqrt{3}+2\sqrt{2} +2} = a+b\sqrt{3}}$

$\implies\rm{\dfrac{1+4\sqrt{3}}{(5-6)\sqrt{3} +(4-6+2)\sqrt{2} +2}}$

$\implies\rm{\dfrac{1+4\sqrt{3}} {2-\sqrt{3}}}$

Now Rationalising it.

$\implies\rm{\dfrac{1+4\sqrt{4}}{2-\sqrt{3}}\times{\dfrac{2+\sqrt{3}}{2+\sqrt{3}}}}$

$\implies\rm{\dfrac{2+\sqrt{3} +8\sqrt{3} +12}{1}}$

$\implies\rm{14+9\sqrt{3} =a +b \sqrt{3}}$

$\implies\rm{14+9\sqrt{\cancel{3}} = a+b\sqrt{\cancel{3}}}$.

Hence, a = 14, b = 9.