Question

The solution set of the equation \[(x+1)(x+2)(x+3)(x+4)=120\] is
A.\[\{-6,1,[-5\pm \sqrt{39}i]/2\}\]
B.\[\{6,-1,[-5\pm \sqrt{39}i]/2\}\]
C.\[\{-6,-1,[-5\pm \sqrt{39}i]/2\}\]
D.\[\{6,1,[-5\pm \sqrt{39}i]/2\}\]

Answer 1

Hint: To solve this type of problem firstly simplify the given equation and bring it in the square form and after that use the Sridharacharya method to obtain the values of variable \[x\] and after calculating all the values you will get the required set of values of the given equation.

Complete step-by-step answer:
Before solving the given problem, first let’s understand the Sridharacharya method:
 Sridhara wrote down rules for solving quadratic equations, it is the most common method of finding the roots of the quadratic equation and it is known as Sridharacharya rule. The quotation of Sridhara’s rule is as: “Multiply both sides of the equation by a known quantity equal to four times the coefficient of the square of the unknown; add to both sides a known quantity equal to the square of the coefficient of the unknown; then take the square root.
Let’s consider a quadratic equation: \[a{{x}^{2}}+bx+c\] , where \[a,b,c\] are real numbers and numeral coefficients. And here \[a\ne 0\] , because if it is equal to zero then the equation will not remain quadratic anymore and it will become a linear equation.
This formula for a quadratic equation is used to find the roots of the equation. Since, quadratic equations have a degree equal to two, hence there will be two solutions for the equation.
The formula to find the roots of this equation will be:
\[x=\dfrac{\left[ -b\pm \sqrt{({{b}^{2}}-4ac)} \right]}{2a}\]
In this, the nature of the roots depends on \[({{b}^{2}}-4ac)\] . This is called as the Discriminant and it is represented as \[D\]
Therefore, Discriminant \[D={{b}^{2}}-4ac\]
There are three cases after finding the value of \[D\]
Case1:- When \[D>0\] , then the roots are real and distinct in nature.
 Case2:- When \[D=0\] , then the roots are real and equal in nature.
Case3:- When \[D<0\] , then the roots are not real.
Now, according to the given question:
\[(x+1)(x+2)(x+3)(x+4)=120\]
\[\Rightarrow ({{x}^{2}}+5x+4)({{x}^{2}}+5x+6)=120\]
Let, \[{{x}^{2}}+5x=m\]
\[\Rightarrow (m+4)(m+6)=120\]
\[\Rightarrow (m+4)(m+6)=10*12\]
\[\begin{align}
  & \Rightarrow (m+4)=10 \\
 & \Rightarrow m=10-4 \\
 & \Rightarrow m=6 \\
\end{align}\] And
 \[\begin{align}
  & \Rightarrow (m+4)=-12 \\
 & \Rightarrow m=-12-4 \\
 & \Rightarrow m=-16 \\
\end{align}\]
Now, \[{{x}^{2}}+5x=6\]
\[\Rightarrow {{x}^{2}}+5x-6=0\]
\[\Rightarrow {{x}^{2}}+(6-1)x-6=0\]
\[\Rightarrow {{x}^{2}}+6x-1x-6=0\]
\[\Rightarrow x(x+6)-1(x+6)=0\]
\[\Rightarrow (x+6)(x-1)=0\]
\[\Rightarrow x=-6,x=1\]
And \[{{x}^{2}}+5x=-16\]
\[\Rightarrow {{x}^{2}}+5x+16=0\]
Now, using Sri Dharacharya method:
 \[d={{(5)}^{2}}-4\times 1\times 16\]
\[d=-39\]
\[\Rightarrow x=\dfrac{-b\pm \sqrt{d}}{2a}\]
\[\Rightarrow x=\dfrac{-5\pm \sqrt{39}i}{2}\]
So, the solution set can be defined as: \[\{-6,1,\dfrac{-5\pm \sqrt{39}i}{2}\}\]
Hence, the correct option from all the above option is \[A\]
So, the correct answer is “Option A”.

Note: A quadratic equation makes a parabola when graphed on a coordinate plane. If the value of Discriminant is positive, then the graph crosses the x-axis, When the value of Discriminant is zero, then the graph touches the x-axis but does not cross it, When the value of Discriminant is negative, then the graph does not even touch the x-axis.