Question

Simplify: ${(1.5p + 1.2q)^2} - {(1.5p - 1.2q)^2}$

Answer 1

Hint: The addition is the sum of given two or more than two numbers, or variables and in addition, if we sum the two or more numbers then we obtain a new frame of the number will be found which can be represented as $ + $, also in subtraction which is the minus of given two or more than two numbers, but here comes with the condition that in subtraction the greater number sign represented in the number will stay constant example $2 - 3 = - 1$
Making use of these operations, the generalized form of algebraic expansion ${(a + b)^2}$ Is found.

Complete step by step solution:
Given the equation ${(1.5p + 1.2q)^2} - {(1.5p - 1.2q)^2}$ and then we need to find the value of the simplified form.
Since we know that ${(a + b)^2} = (a + b)(a + b) \Rightarrow {a^2} + {b^2} + 2ab$ and also ${(a - b)^2} = (a - b)(a - b) \Rightarrow {a^2} + {b^2} - 2ab$
Thus, making use of these formulas we get \[{(1.5p + 1.2q)^2} = (1.5p + 1.2q) \times (1.5p + 1.2q) = (2.25{p^2} + 1.44{q^2} + 3.6pq)\]
\[{(1.5p - 1.2q)^2} = (1.5p - 1.2q) \times (1.5p - 1.2q) = (2.25{p^2} + 1.44{q^2} - 3.6pq)\]
Hence using the subtraction operation, we get \[(2.25{p^2} + 1.44{q^2} + 3.6pq) - (2.25{p^2} + 1.44{q^2} - 3.6pq)\]
Further solving with giving the bracket inside the negative terms we get, \[(2.25{p^2} + 1.44{q^2} + 3.6pq) - 2.25{p^2} - 1.44{q^2} + 3.6pq\]
Hence canceling the common terms, we get \[(2.25{p^2} + 1.44{q^2} + 3.6pq) - 2.25{p^2} - 1.44{q^2} + 3.6pq = 3.6pq + 3.6pq\]
Thus, using the addition operation, we get ${(1.5p + 1.2q)^2} - {(1.5p - 1.2q)^2} = 7.2pq$

Note: Things that we need to keep in mind while solving these types of questions that is simplifying the algebraic equation are: we first need to simplify the exponents so that we will end up getting algebraic equation in simpler form then we need to be careful in grouping like terms because we are not supposed to group unlike terms. Also, we have to be more careful while simplifying the expressions after expanding them using the formulae. There may be a chance of making mistakes in that.