Question

What is the simplest form of radical expression of $\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}}$?

Answer 1

Hint: The simplest form of such an expression never contains irrationals in the denominator, so we need to think of a trick such that the irrationality from the denominator is eliminated. Without caring about what the numerator might turn out to be. And to rationalise an irrational we simply use the identity $a^2-b^2=(a+b)(a-b)$ in the denominator which is what we are going to use here .

Complete step-by-step answer:
We have the expression:
$\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}}$
Multiplying both numerator and denominator by $\sqrt2+\sqrt5$, we get:
$\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}}\times\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}$
Using the identity $a^2-b^2=(a+b)(a-b)$ in denominator, and making square out of the terms present in the numerator, we get the expression:
$\dfrac{{{(\sqrt{2}+\sqrt{5})}^{2}}}{{{(\sqrt{2})}^{2}}-{{(\sqrt{5})}^{2}}}$
Further simplifying it, and using the identity $(a+b)^2=a^2+b^2+2ab$ in the numerator:
$\dfrac{2+5+(2\times\sqrt{2}\times\sqrt{5})}{2-5}=\dfrac{7+2\sqrt{10}}{-3}$
And now we can split the rational and irrational for the simplest form, so we can write it as:
$\dfrac{-7}{3}+\dfrac{-2\sqrt{10}}{3}$
Which is the simplest form of the given radical expression.
Hence, $\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}}$=$\dfrac{-7}{3}+\dfrac{-2\sqrt{10}}{3}$

Note: Take care of the signs while simplifying the expression. Use the identities correctly, and multiply both numerator and denominator by the same expression, anything done otherwise is not valid. If you try to create a squared term in the denominator, you will again be stuck with an irrational there, so always try to create $a^2-b^2$ type terms in the denominator so that elimination of the irrational is easy from there.