Question

What is the rationalising factor of the given number? $$2+\sqrt{7+2\sqrt{10}}$$

Answer 1

Hint: In order to solve this question, first we will simplify the expression $$\sqrt{7+2\sqrt{10}}$$ . For this we will use the concept that if a given expression is of the form, $$\sqrt{p+q\sqrt{r}}$$ and $$p^2-q^{2}r$$ is a perfect square $$s^2$$ then $$\sqrt{p+q\sqrt{r}}={\displaystyle \frac{\sqrt{2p+2s}}{2}}+{\displaystyle \frac{\sqrt{2p-2s}}{2}}$$ .We will substitute the values in the formula and simplify the expression. After that we will use the definition of rationalising factor to find the rationalising factor of the given number. The rationalising factor of a given number is the number by which it must be multiplied so that it becomes rational.
Complete step by step answer:
Given: $$2+\sqrt{7+2\sqrt{10}}$$
Now, first of all we will simplify the expression $$\sqrt{7+2\sqrt{10}}$$
We know that
if a given expression is of the form, $$\sqrt{p+q\sqrt{r}}$$ and $$p^2-q^{2}r$$ is a perfect square $$s^2$$
then $$\sqrt{p+q\sqrt{r}}={\displaystyle \frac{\sqrt{2p+2s}}{2}}+{\displaystyle \frac{\sqrt{2p-2s}}{2}}$$
In our problem, $$p=7,q=2,r=10$$
So, $$p^2-q^{2}r=7^2-2^2\cdot \left(10\right)$$
$$\Rightarrow p^2-q^{2}r=49-40$$
$$\Rightarrow p^2-q^{2}r=9=s^2$$
Taking square root, we get
$$s=3$$
Thus, we get $$p^2-q^{2}r$$ is a perfect square $$s^2$$
It means $$\sqrt{p+q\sqrt{r}}={\displaystyle \frac{\sqrt{2p+2s}}{2}}+{\displaystyle \frac{\sqrt{2p-2s}}{2}}$$
Therefore, on substituting the values, we get
$$\sqrt{7+2\sqrt{10}}={\displaystyle \frac{\sqrt{2\left(7\right)+2\left(3\right)}}{2}}+{\displaystyle \frac{\sqrt{2\left(7\right)-2\left(3\right)}}{2}}$$
$$\Rightarrow \sqrt{7+2\sqrt{10}}={\displaystyle \frac{\sqrt{14+6}}{2}}+{\displaystyle \frac{\sqrt{14-6}}{2}}$$
On adding the terms in the numerator, we get
$$\Rightarrow \sqrt{7+2\sqrt{10}}={\displaystyle \frac{\sqrt{20}}{2}}+{\displaystyle \frac{\sqrt{8}}{2}}$$
We can write
$$\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}$$
$$\sqrt{8}=\sqrt{2\times 2\times 2}=2\sqrt{2}$$
Therefore, above equation becomes,
$$\Rightarrow \sqrt{7+2\sqrt{10}}={\displaystyle \frac{2\sqrt{5}}{2}}+{\displaystyle \frac{2\sqrt{2}}{2}}$$
On cancelling $$2$$ from both numerator and denominator, we get
$$\Rightarrow \sqrt{7+2\sqrt{10}}=\sqrt{5}+\sqrt{2}$$
Hence,
$$2+\sqrt{7+2\sqrt{10}}=2+\sqrt{5}+\sqrt{2}$$
Now we know that
Rationalising factor of a given number is the number by which it must be multiplied so that it becomes rational.
So, we have to find a number by which the number is multiplied so that it becomes rational.
Since, the number involves two square roots, a suitable radical conjugate is formed by multiplying the variants of the expression $$2\pm \sqrt{5}\pm \sqrt{2}$$ apart from the one we already have
i.e., $$\left(2+\sqrt{5}-\sqrt{2}\right)\left(2-\sqrt{5}-\sqrt{2}\right)\left(2-\sqrt{5}+\sqrt{2}\right)$$
we know that
$$\left(a+b\right)\left(a-b\right)=a^2-b^2$$
Therefore, from the last two terms, we get
$$=\left(2+\sqrt{5}-\sqrt{2}\right)\left({\left(2-\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2\right)$$
Now we know that
$${\left(a-b\right)}^2=a^2+b^2-2ab$$
Therefore, we get
$$=\left(2+\sqrt{5}-\sqrt{2}\right)\left(4+5-4\sqrt{5}-2\right)$$
$$=\left(2+\sqrt{5}-\sqrt{2}\right)\left(7-4\sqrt{5}\right)$$
On multiplying we get
$$=7\left(2+\sqrt{5}-\sqrt{2}\right)-4\sqrt{5}\left(2+\sqrt{5}-\sqrt{2}\right)$$
$$=14+7\sqrt{5}-7\sqrt{2}-8\sqrt{5}-20+4\sqrt{10}$$
$$=-6-\sqrt{5}-7\sqrt{2}+4\sqrt{10}$$
Note that this is negative, so let’s negate it to get the slightly more attractive radical conjugate
Therefore, we get
$$=6+\sqrt{5}+7\sqrt{2}-4\sqrt{10}$$
which is the required rationalising factor of $$2+\sqrt{7+2\sqrt{10}}$$

Note:
 Students should always remember that the rationalising factor should be the smallest number which when multiplied by the given number gives a rational number.
As a check let’s multiply $$6+\sqrt{5}+7\sqrt{2}-4\sqrt{10}$$ and $$2+\sqrt{5}+\sqrt{2}$$
i.e., $$\left(6+\sqrt{5}+7\sqrt{2}-4\sqrt{10}\right)\left(2+\sqrt{5}+\sqrt{2}\right)$$
$$=2\left(6+\sqrt{5}+7\sqrt{2}-4\sqrt{10}\right)+\sqrt{5}\left(6+\sqrt{5}+7\sqrt{2}-4\sqrt{10}\right)+\sqrt{2}\left(6+\sqrt{5}+7\sqrt{2}-4\sqrt{10}\right)$$
$$=\left(12+2\sqrt{5}+14\sqrt{2}-8\sqrt{10}\right)+\left(6\sqrt{5}+5+7\sqrt{10}-20\sqrt{2}\right)+\left(6\sqrt{2}+\sqrt{10}+14-8\sqrt{5}\right)$$
$$=31$$ which is a rational number
Hence, our answer is correct.