Question

How do you solve \[{\left( {x + a} \right)^2} - {b^2} = 0\] by factoring?

Answer 1

Hint: We can solve this using simple algebraic identity. Here we apply \[{(p + q)^2}\] formula that is \[{(p + q)^2} = {p^2} + 2pq + {q^2}\] and simplifying we will have a quadratic equation. We can factor this by splitting the middle terms. On further simplifying we will have a desired result.

Complete step-by-step answer:
Given,
 \[{\left( {x + a} \right)^2} - {b^2} = 0\]
Now applying the identity \[{(p + q)^2} = {p^2} + 2pq + {q^2}\] ,
 \[{x^2} + {a^2} + 2ax - {b^2} = 0\]
 \[{x^2} + 2ax + {a^2} - {b^2} = 0\]
We can split 2a as \[2a = \left( {a + b} \right) + \left( {a - b} \right)\] then we have,
 \[{x^2} + \left( {(a + b) + (a - b)} \right)x + {a^2} - {b^2} = 0\]
We can write \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
 \[{x^2} + (a + b)x + (a - b)x + (a + b)(a - b) = 0\]
Now for the first two terms we take x common and for the remaining two terms we take \[\left( {a - b} \right)\] common
 \[x(x + (a + b)) + (a - b)(x + (a + b)) = 0\]
Now taking \[(x + (a + b))\] common we will have
 \[(x + (a + b))(x + (a - b)) = 0\]
now from the zero product principle we will have,
 \[ \Rightarrow (x + (a + b)) = 0\] and \[(x + (a - b)) = 0\]
 \[ \Rightarrow x = - \left( {a + b} \right)\] and \[x = - \left( {a - b} \right)\]
This is the required solution of \[{\left( {x + a} \right)^2} - {b^2} = 0\]
So, the correct answer is “ \[{\left( {x + a} \right)^2} - {b^2} = 0\] ”.

Note: We can solve this using simple algebraic identity. That is if we observe the given problem it is of the form \[{p^2} - {q^2}\] where \[p = x + a\] and \[q = b\] .
 \[{\left( {x + a} \right)^2} - {b^2} = 0\]
Then now applying the identity \[{p^2} - {q^2} = \left( {p + q} \right)\left( {p - q} \right)\]
 \[\left( {\left( {x + a} \right) + b} \right)\left( {\left( {x + a} \right) - b} \right) = 0\]
By zero product principle we have,
 \[\left( {\left( {x + a} \right) + b} \right) = 0\] and \[\left( {\left( {x + a} \right) - b} \right) = 0\]
 \[ \Rightarrow x = - \left( {a + b} \right)\] and \[x = - (a - b)\] . In both the cases we have the same answer. If we have a polynomial of degree n then we will have n roots or n zeros. Here we have a polynomial of degree 2 hence we have two roots.