gps5441 wrote:
pushpitkc wrote:
Let X = \((1+ \sqrt{3})\sqrt{(2+3^{1/2})}\) = \((1+ \sqrt{3})\sqrt{(1 + 1 + 3^{1/2})}\)
If a = \(1+ \sqrt{3}\), then \(a^2 = 1 + 3 + 2\sqrt{3} = 4 + 2\sqrt{3}\)
The expression now becomes X = \(a * \sqrt{1 + a}\)
Squaring
\(X^2 = a^2(1+a)\)
Substituting values,
\(X^2 = (4 + 2\sqrt{3})(2 + \sqrt{3})\)
\(= 2(2 + \sqrt{3})(2 + \sqrt{3})\)
Taking square root
X = \(\sqrt{2}(2 + \sqrt{3})\) (Option A)
Are we allowed to take 2 common from one bracket set?
eg 6X7=2(3)(7)
Yes you can take '2' common from one braket.
Notice that when you multiply '2' back to the braket, you'll get the previous equation.
I'll explain with example.
Case 1: Let us say an equation is x = (2y + 4)*(y^2 + 3y); here x and y are some variables.
In this case you can take '2' common from first bracket and 'y' common from second bracket. Equation can be written as, x = 2y*(y+2)*(y+3).
You can also multiply '2' and 'y' separately to the brackets, even to the bracket from which each was not taken.
Equation can also be written as, x = (y^2+2y)*(2y+6) = 2*(y^2+2y)*(y+3) = y*(y+2)*(2y+6). All these equations are same and correct.
Case 2: Let us say an equation is x = (2y + 4) + (y^2 + 3y); here x and y are some variables.
In this case also you can take '2' common from first bracket and 'y' common from second bracket but you cannot multiply both commons like in previous case.
Equation will be written as, x = 2(y+2) + y(y+3). This equation is same as original equation.
Case 3: Let us say an equation is x = (3y + 6) + (y^2 + 2y); here x and y are some variables.
In this case also you can take '2' common from first bracket and 'y' common from second bracket. Equation can be written as, x = 3(y+2)+y(y+2).
In this case, you can take '(y+2)' as common and equation will become, x = (y+2)*(y+3).
I hope this clears your doubt.