Question

Find the square root of $\left( {{x^2} + 4x + 4} \right)\left( {{x^2} + 6x + 9} \right)$.

Answer 1

Hint: We will first write the given function as a product of two different functions. Then, we will find the factors for both of the functions using the square method, we will see that these are square of some factors and thus our square root will get cancelled out.

Complete step by step answer:
Let us say, we have $f(x) = \left( {{x^2} + 4x + 4} \right)\left( {{x^2} + 6x + 9} \right)$.
So, we basically need to find $\sqrt {f(x)} = \sqrt {\left( {{x^2} + 4x + 4} \right)\left( {{x^2} + 6x + 9} \right)} $ ……….(1)
Assume that it is the product of two different function g(x) and h(x) such that $g(x) = {x^2} + 4x + 4$ and $h(x) = {x^2} + 6x + 9$
Now, let us first consider $g(x)$.
$ \Rightarrow g(x) = {x^2} + 4x + 4$
We can write this as
$ \Rightarrow g(x) = {\left( x \right)^2} + 2 \times \left( 2 \right) \times \left( x \right) + {\left( 2 \right)^2}$ ……………(2)
Now, we know that we have a formula
${(a + b)^2} = {a^2} + {b^2} + 2ab$ ………….(3)
Therefore, putting $a = x$ and $b = 2$ in (2), we will get
$ \Rightarrow g(x) = {\left( {x + 2} \right)^2}$
Therefore, we get
$ \Rightarrow {x^2} + 4x + 4 = {\left( {x + 2} \right)^2}$ ………….(4)
Now, let us first consider $h(x)$.
$ \Rightarrow h(x) = {x^2} + 6x + 9$
We can write this as
$ \Rightarrow h(x) = {\left( x \right)^2} + 2 \times \left( 3 \right) \times \left( x \right) + {\left( 3 \right)^2}$ ……………(5)
Using the formula written in the equation (3) and putting $a = x$ and $b = 3$ in (5), we will get
$ \Rightarrow h(x) = {\left( {x + 3} \right)^2}$
Therefore, we get
$ \Rightarrow {x^2} + 6x + 9 = {\left( {x + 3} \right)^2}$ ………….(6)
Putting (4) and (6) in (1), we will get
$ \Rightarrow \sqrt {f(x)} = \sqrt {{{\left( {x + 2} \right)}^2}{{\left( {x + 3} \right)}^2}} $
Hence, on simplifying the RHS, we will get
$ \Rightarrow \sqrt {f(x)} = \left( {x + 2} \right)\left( {x + 3} \right)$.

$\therefore$ The square root of $\left( {{x^2} + 4x + 4} \right)\left( {{x^2} + 6x + 9} \right)$ is $ \sqrt {f(x)} = \left( {x + 2} \right)\left( {x + 3} \right)$.

Note:
The students must note and keep in mind that, if they multiply the terms inside the square root, we will get an equation of degree 4 which will be extremely difficult to be factorized or anything. Therefore, you should always try to first work with quadratic equations only, because we have a lot of ways to factorize them, like splitting the middle term or completing the square.
The students must wonder how the identity or formula that we use did come in the picture. We know the binomial theorem which states that ${(a + b)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^n}{b^1} + .......{ + ^n}{C_{n - 1}}{a^1}{b^{n - 1}}{ + ^n}{C_n}{a^0}{b^n}$
Now, put n = 2, we will get:-
${(a + b)^2}{ = ^2}{C_0}{a^2}{b^0}{ + ^2}{C_1}{a^1}{b^1}{ + ^2}{C_2}{a^0}{b^2}$ …….(A)
Now, we will use $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
So, we will get:-
$^2{C_0} = \dfrac{{2!}}{{0!(2)!}} = 1{,^2}{C_1} = \dfrac{{2!}}{{1!(1)!}} = 2{,^2}{C_2} = \dfrac{{2!}}{{2!(0)!}} = 1$
Putting all these in (A), we will get:-
${(a + b)^2} = {a^2} + 2ab + {b^2}$
Thus, we got our formula.