Question

Square root of $ \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{4{x^2}}} - \dfrac{x}{y} + \dfrac{y}{{2x}} - \dfrac{3}{4} $ is $ \dfrac{x}{y} - \dfrac{1}{2} - \dfrac{y}{{2x}} $
A. True
B. False

Answer 1

Hint: As we know that the square root of any number is that number which when multiplied by itself gives the product equal to the number whose square root is determined. Square root of any number is represented within the symbol $ \sqrt {} $ . In this question we will first break down the number and turn into the form of algebraic identities.

Complete step by step solution:
Let us take the left hand side or solve the first part i.e. cc
We can write $ \dfrac{{{x^2}}}{{{y^2}}} = \left( { - \dfrac{x}{y}} \right) $ as when square of negative gives positive, now $ \dfrac{{{y^2}}}{{4{x^2}}} $ can be written as the square of $ {\left( {\dfrac{y}{{2x}}} \right)^2} $ .
Similarly we can write $ \dfrac{{ - 3}}{4} $ as the sums of $ - 1 + \dfrac{1}{4} $ .
Writing them all together we have $ {\left( { - \dfrac{x}{y}} \right)^2} + {\left( {\dfrac{y}{{2x}}} \right)^2} - \dfrac{x}{y} + \dfrac{y}{{2x}} - 1 + \dfrac{1}{4} $ .
We know the algebraic identities i.e. $ {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac $ ,
By applying this in the above expression we have $ $ $ {\left( { - \dfrac{x}{y}} \right)^2} + {\left( {\dfrac{y}{{2x}}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^2} + \left\{ {2 \times \left( { - \dfrac{x}{y}} \right)\left( {\dfrac{y}{{2x}}} \right)} \right\} + \left\{ {2 \times \left( {\dfrac{y}{{2x}}} \right)\left( {\dfrac{1}{2}} \right)} \right\} + \left\{ {2 \times \left( {\dfrac{1}{2}} \right)\left( {\dfrac{{ - x}}{y}} \right)} \right\} $ .
So they can be shortly written as $ {\left( { - \dfrac{x}{y} + \dfrac{y}{{2x}} + \dfrac{1}{2}} \right)^2} $ .
Now we will find the square root of $ \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{4{x^2}}} - \dfrac{x}{y} + \dfrac{y}{{2x}} - \dfrac{3}{4} $ as follows: $ \sqrt {\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{4{x^2}}} - \dfrac{x}{y} + \dfrac{y}{{2x}} - \dfrac{3}{4}} \; = \sqrt {{{\left( {\dfrac{{ - x}}{y} + \dfrac{y}{{2x}} + \dfrac{1}{2}} \right)}^2}} $ .
It gives us the value $ \pm \left( { - \dfrac{x}{y} + \dfrac{y}{{2x}} + \dfrac{1}{2}} \right) $ .
Hence the correct option is (A) True
So, the correct answer is “Option A”.

Note: Before solving this kind of question we should have the full knowledge of algebraic identities and their formulas. We should also note that the value of any square root is both positive and negative value, it is commonly mistaken to forget the negative value of the sign $ \pm $ . It means that there are two values of any number when it is written under square root.