Question

What is one of the square roots of $$9-2\sqrt{14}$$

(A). $$\sqrt{7} -\sqrt{3}$$

(B). $$\sqrt{6} -\sqrt{3}$$

(C). $$\sqrt{7} -\sqrt{5}$$

(D). $$\sqrt{7} -\sqrt{2}$$

Answer 1

Hint: In this question it is given that we have to find one of the square roots of $$9-2\sqrt{14}$$. So for this we have to consider the given expression as $$\left( a-b\right)^{2} $$, from which by solving we have to find the value of a and b and the result of a+b gives our required solution.

Complete step-by-step solution:

Given expression, $$9-2\sqrt{14}$$

Let us consider that $$\left( a+b\right)^{2} =9-2\sqrt{14}$$

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,

$$a^{2}+2ab+b^{2}=9-2\sqrt{14}$$

$$\Rightarrow a^{2}+b^{2}+2ab=9-2\sqrt{14}$$.........(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,

$$a^{2}+b^{2}=9$$.......(2) and $$2ab=-2\sqrt{14}$$............(3)

From equation (3) we can write,

$$b=\dfrac{2\sqrt{14}}{-2a}$$

$$\Rightarrow b=-\dfrac{\sqrt{14} }{a}$$..................(4)

Now putting the value of b in equation (2), we get,

$$a^{2}+\left( -\dfrac{\sqrt{14} }{a} \right)^{2} =9$$

$$\Rightarrow a^{2}+\dfrac{14}{a^{2}} =9$$ ($$\because \left( \dfrac{a}{b} \right)^{n} =\dfrac{a^{n}}{b^{n}}$$)

Let $$a^{2}=x$$, then the above equation can be written as,

$$x+\dfrac{14}{x} =9$$

$$\Rightarrow x^{2}+14=9x$$ ( multiplying both side by x)

$$\Rightarrow x^{2}-9x+14=0$$......(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=-9, c=14), we get,

$$x=\dfrac{-\left( -9\right) \pm \sqrt{\left( -9\right)^{2} -4\times 1\times 14} }{2\times 1}$$

$$=\dfrac{9\pm \sqrt{81-56} }{2\times 1}$$

$$=\dfrac{9\pm \sqrt{25} }{2}$$

$$=\dfrac{9\pm5}{2}$$

$$\text{Either} \ x=\dfrac{9+5}{2} \ \text{or} \ x=\dfrac{9-5}{2}$$

$$\Rightarrow \ x=\dfrac{14}{2} \ \text{or} \ x=\dfrac{4}{2}$$

$$\Rightarrow \ x=7\ \text{or} \ x=2$$

$$\Rightarrow \ a^{2}=7\ \text{or} \ a^{2}=2$$

$$\Rightarrow \ a=\pm \sqrt{7} \ \text{or} \ a=\pm \sqrt{2}$$

Now equation (3) also can be written as,

$$b=-\dfrac{\sqrt{ 2\times 7} }{a}$$

$$b=-\dfrac{\sqrt{2} \times \sqrt{7} }{a}$$......(5) [

Now putting the values of a in equation (5), we get several solutions,

If $$a=\sqrt{7}$$, then $$b=-\dfrac{\sqrt{2} \times \sqrt{7} }{\sqrt{7} } =-\sqrt{2}$$

$$\therefore \left( a+b\right) =\left( \sqrt{7} -\sqrt{2} \right) $$

If $$a=-\sqrt{7}$$, then $$b=-\dfrac{\sqrt{2} \times \sqrt{7} }{(-\sqrt{7}) } =\sqrt{2}$$

$$\therefore \left( a+b\right) =-\left( \sqrt{7} -\sqrt{2} \right) $$,

Similarly for $$a=\sqrt{2}$$ and $$a=-\sqrt{2}$$ we will get the solution of (a+b)=$$-\left( \sqrt{7} -\sqrt{2} \right) $$ and (a+b)=$$\left( \sqrt{7} -\sqrt{2} \right) $$

Therefore, we can say that the possible values of $$\left( a+b\right) is \left( \sqrt{7} - \sqrt{2} \right) $$ or $$-\left( \sqrt{7} - \sqrt{2} \right) $$

But according to the opposition the matching option is option D.

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 $$\left( a+b\right)^{2} =9+2\sqrt{14}$$, which gives, $$ a+b=\pm \sqrt{9+2\sqrt{14} }$$, 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 $$2\sqrt{14}$$ 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.