Question

The positive square root of $$11+\sqrt{112}$$ is:
(A). $$\sqrt{7} +2$$
(B). $$\sqrt{7} +\sqrt{2}$$
(C). $$2-\sqrt{7}$$
(D). $$\sqrt{7} -\sqrt{2}$$

Answer 1

Hint: In this question it is given that we have to find the positive square root of $$11+\sqrt{112}$$. So for this we have to consider the given expression as $$\left( a+b\right)^{2} $$, from where by solving we have to find the value of a and b and the positive value of a+b gives our required solution.

Complete step-by-step solution:
Given expression, $$11+\sqrt{112}$$
Let us consider that $$\left( a+b\right)^{2} =11+\sqrt{112}$$
Where a and b are any two real numbers.
Now as we know that $$\left( a+b\right)^{2} =a^{2}+2ab+b^{2}$$, so by using the identity we can write the above equation as,
$$a^{2}+2ab+b^{2}=11+\sqrt{112}$$
$$\Rightarrow a^{2}+b^{2}+2ab=11+\sqrt{112}$$.........(1)
As we know that the square of any rational or irrational number gives a rational number then $$a^{2}$$ and $$b^{2}$$ must be a rational number.
since sum of two rational number gives a rational number, therefore, $$a^{2}+b^{2}$$ is also a rational number,
Therefore form equation (1) we can write,
$$a^{2}+b^{2}=11$$.......(2) and $$2ab=\sqrt{112}$$............(3)
From equation (3) we can write,
$$b=\dfrac{\sqrt{112} }{2a}$$
$$\Rightarrow b=\dfrac{\sqrt{2\times 2\times 28} }{2a}$$
$$\Rightarrow b=\dfrac{2\sqrt{28} }{2a}$$ [$$\because \sqrt{a\times a} =a$$]
$$\Rightarrow b=\dfrac{\sqrt{28} }{a}$$
Now putting the value of b in equation (2), we get,
$$a^{2}+\left( \dfrac{\sqrt{28} }{a} \right)^{2} =11$$
$$\Rightarrow a^{2}+\dfrac{28}{a^{2}} =11$$
Let $$a^{2}=x$$, then the above equation can be written as,
$$x+\dfrac{28}{x} =11$$
$$\Rightarrow x^{2}+28=11x$$ [ multiplying both side by x]
$$\Rightarrow x^{2}-11x+28=0$$......(4)
Now as we know that if any equation is in the form of $$ax^{2}+bx+c=0$$,
Then the solution, $$x=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$$
By using the above quadratic formula (where a=1, b=-11, c=28), we get,
$$x=\dfrac{-\left( -11\right) \pm \sqrt{\left( -11\right)^{2} -4\times 1\times 28} }{2\times 1}$$
  $$=\dfrac{11\pm \sqrt{121-112} }{2}$$
  $$=\dfrac{11\pm \sqrt{9} }{2}$$
  $$=\dfrac{11\pm 3}{2}$$
$$\text{Either} \ x=\dfrac{11+3}{2} \ \text{or} \ x=\dfrac{11-3}{2}$$
         $$\Rightarrow \ x=\dfrac{14}{2} \ \text{or} \ x=\dfrac{8}{2}$$
         $$\Rightarrow \ x=7\ \text{or} \ x=4$$
         $$\Rightarrow \ a^{2}=7\ \text{or} \ a^{2}=4$$
         $$\Rightarrow \ a=\pm \sqrt{7} \ \text{or} \ a=\pm \sqrt{4}$$
         $$\Rightarrow \ a=\pm \sqrt{7} \ \text{or} \ a=\pm 2$$
Now equation (3) also can be written as,
$$b=\dfrac{\sqrt{2\times 2\times 7} }{a}$$
$$\Rightarrow b=\dfrac{2\sqrt{7} }{a}$$.......(5) [$$\because \sqrt{a\times a} =a$$]
Now putting the values of a in equation (5), we get several solutions,
If $$a=\sqrt{7}$$, then $$b=\dfrac{2\sqrt{7} }{\sqrt{7} } =2$$
$$\therefore \left( a+b\right) =\left( \sqrt{7} +2\right) $$
If $$a=-\sqrt{7}$$, then $$b=\dfrac{2\sqrt{7} }{(-\sqrt{7}) } =-2$$
$$\therefore \left( a+b\right) =-\left( \sqrt{7} +2\right) $$, which is not possible( since the result is positive)
Similarly for a=+2 and a=-2 we will get the solution of (a+b)=$$\left( \sqrt{7} +2\right) $$ and (a+b)=$$-\left( \sqrt{7} +2\right) $$
Therefore, we can say that the positive value of $$\left( a+b\right) =\left( \sqrt{7} +2\right) $$
Hence the correct option is option A.

Note: The doubt may arise that while solving, we can directly find the value of (a+b) just by square root in both side of the equation $$\left( a+b\right)^{2} =11+\sqrt{112}$$, which gives, $$\Rightarrow a+b=\pm \sqrt{11+\sqrt{112} }$$, but instead of that why we follow the above lengthy process in order to get the value of (a+b), this is because in the above mentioned step this $$\sqrt{112}$$ is already in irrational form of irrational number, so if you again do square root of this term then this will not lead to the desired solution.