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One root of the equation $\left( {x + 1} \right)\left( {x + 3} \right)\left( {x + 2} \right)\left( {x + 4} \right) = 120$ is
A). $ - 1$
B). $2$
C). $1$
D). $0$

seo-qna
Last updated date: 21st Jul 2024
Total views: 347.4k
Views today: 7.47k
Answer
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Hint: The root of an equation is nothing but the value which satisfies the equation when substituted for an unknown quantity in an equation.
An equation can contain more than one root.
Formula used:
The quadratic formula is as follows.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where,$a,b,c$ are constants and$a \ne 0$
$x$ is an unknown quantity.

Complete step-by-step solution:
 We need to solve the given equation $\left( {x + 1} \right)\left( {x + 3} \right)\left( {x + 2} \right)\left( {x + 4} \right) = 120$.
 Let us rearrange the given equation as $\left( {x + 1} \right)\left( {x + 4} \right)\left( {x + 2} \right)\left( {x + 3} \right) = 120$ for our convenience.
 We shall combine the first two equations and last two equations.
 We get, \[({x^2} + 5x + 4)({x^2} + 5x + 6) = 120\]
Let us assume ${x^2} + 5x = y$ so that we can solve the given equation easily.
So we get, \[(y + 4)(y + 6) = 120\]
Combining the left hand side, we get,
 \[{y^2} + 10y + 24 = 120\]
\[\Rightarrow {y^2} + 10y + 24 - 120 = 0\]
$ \Rightarrow {y^2} + 10y - 96 = 0$
\[ \Rightarrow {y^2} + 16y - 6y - 96 = 0\] (Expressing $10y$ as $16y - 6y$)
\[ \Rightarrow y(y + 16) - 6(y + 16) = 0\] (We have taken common $y$ terms)
Further solving, we have
$ \Rightarrow (y + 16)(y - 6) = 0$
Therefore, the roots of y are calculated as:
$\begin{gathered}
  y + 16 = 0 \\
   \Rightarrow y = - 16 \\
\end{gathered} $
$\begin{gathered}
  y - 6 = 0 \\
   \Rightarrow y = 6 \\
\end{gathered} $
Since we have assumed ${x^2} + 5x = y$, we need to substitute the values of $y$ here.
Hence, ${x^2} + 5x = - 16$
\[ \Rightarrow {x^2} + 5x + 16 = 0\]
Use a quadratic formula to solve the above equation.
Here, $a = 1$,$b = 5$ and $c = - 16$
$ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {{5^2} - 4 \times 1 \times 16} }}{{2 \times 1}}$
$ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 - 64} }}{2}$
\[\Rightarrow x = \dfrac{{ - 5 \pm \sqrt { - 39} }}{2}\]
\[\Rightarrow x = \dfrac{{ - 5}}{2} \pm \dfrac{{\sqrt { - 39} }}{2}\]
\[\Rightarrow x = \dfrac{{ - 5}}{2} \pm \dfrac{{\sqrt {39} }}{2}i\] (Here $i$ is an imaginary complex number)
Hence, we get the following two complex solutions.
\[x = \dfrac{{ - 5}}{2} + \dfrac{{\sqrt {39} }}{2}i\]
And
\[x = \dfrac{{ - 5}}{2} - \dfrac{{\sqrt {39} }}{2}i\]
Now, consider${x^2} + 5x = 6$.
\[ \Rightarrow {x^2} + 5x - 6 = 0\]
\[ \Rightarrow {x^2} + 6x - x - 6 = 0\] (Expressing $5x$ as $6x - x$)
\[ \Rightarrow x(x + 6) - 1(x + 6) = 0\] (We have taken common $y$ terms)
$ \Rightarrow (x + 6)(x - 1) = 0$
Therefore, the roots of$x$ are calculated as:
$\begin{gathered}
  x + 6 = 0 \\
   \Rightarrow x = - 6 \\
\end{gathered} $
$\begin{gathered}
  x - 1 = 0 \\
\Rightarrow x = 1 \\
\end{gathered} $
 Hence, the four roots of the given equation $\left( {x + 1} \right)\left( {x + 3} \right)\left( {x + 2} \right)\left( {x + 4} \right) = 120$ are as follows. \[\;\;x = \dfrac{{ - 5}}{2} + \dfrac{{\sqrt {39} }}{2}i\]
\[ x = \dfrac{{ - 5}}{2} - \dfrac{{\sqrt {39} }}{2}i\]
$x = - 6$
$x = 1$
Here only x=-1 matches with our option. Hence, option C is correct.

Note: Simplification of an expression is the process of changing the expression in an effective manner without changing the meaning of an expression.
Moreover, there are various steps that are involved to simplify an algebraic expression. Some of the steps are listed below:
$>$ If the given algebraic expression contains like terms, we need to combine them.
Example: $3x + 2x + 4 = 5x + 4$
$>$ We need to split an algebraic expression into factors (i.e) the process of finding the factors for the given expression.
Example: ${x^2} + 4x + 3 = (x + 3)(x + 1)$
$>$ We need to expand an algebraic expression (i.e) we have to remove the respective brackets of an expression.
Example: $3(a + b) = 3a + 3b$.
$>$ We need to cancel out the common terms in an algebraic expression.
Example: $\dfrac{{{x^2} + 4x + 3}}{{x + 1}} = \dfrac{{(x + 3)(x + 1)}}{{x + 1}}= x + 3$