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# Evaluate ${\left( {\dfrac{a}{{2b}} + \dfrac{{2b}}{a}} \right)^2} - {\left( {\dfrac{a}{{2b}} - \dfrac{{2b}}{a}} \right)^2} - 4$A) 2B) abC) aD) 0

Last updated date: 13th Jul 2024
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Answer
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Hint: Students if we observe the problem, it is somewhat like ${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} - 4$. This is the hint to solve the question. Instead of solving the original question we will consider the terms as a and b. and then solve it using the identity. And the answer of the identity will be replaced by the considered values.

Complete step by step solution:
Given that,
${\left( {\dfrac{a}{{2b}} + \dfrac{{2b}}{a}} \right)^2} - {\left( {\dfrac{a}{{2b}} - \dfrac{{2b}}{a}} \right)^2} - 4$
Now if we compare it is same as ${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} - 4$
We know that,
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
So let a=$\dfrac{a}{{2b}}$ and b=$\dfrac{{2b}}{a}$ . now we will expand the bracket or equation we substituted.
$= {\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} - 4$
Taking the expansion, we get,
$= {a^2} + 2ab + {b^2} - \left( {{a^2} - 2ab + {b^2}} \right) - 4$
Multiplying the terms by the minus sign we get,
$= {a^2} + 2ab + {b^2} - {a^2} + 2ab - {b^2} - 4$
Now cancel the same terms but with different signs,
$= 2ab + 2ab - 4$
On adding we get,
$= 4ab - 4$
Now we will resubstitute the values of a and b,
$= 4 \times \dfrac{a}{{2b}} \times \dfrac{{2b}}{a} - 4$
Cancelling the same terms in the product,
$= 4 - 4$
When a number is subtracted from the same number the answer is zero.
$= 0$

Therefore, ${\left( {\dfrac{a}{{2b}} + \dfrac{{2b}}{a}} \right)^2} - {\left( {\dfrac{a}{{2b}} - \dfrac{{2b}}{a}} \right)^2} - 4 = 0$. So, option (D) is the correct answer.

Note:
Note that we can also expand the brackets directly but that can be tedious work. So we used the algebraic identities of expansion.
Second point that is to be noted is applicable in general mathematics as, in addition and subtraction we cancel the terms if they are of same value but different signs and the answer is zero; but in multiplication we cancel them when they are same but are present in numerator and denominator patterns regardless of the pattern and the answer is 1 for that cancellation.