Question

The solution set of the equation $(x+1)(x+2)(x+3)(x+4)=120$ is

• A. $\left\{-6,1,\frac{\left[-5\pm \sqrt{39}i\right]}{2}\right\}$
• B. $\left\{6,-1,\frac{\left[-5\pm \sqrt{39i}\right]}{2}\right\}$
• C. $\left\{-6,-1,\frac{\left[-5\pm \sqrt{39i}\right]}{2}\right\}$
• D. $\left\{6,1,\frac{\left[-5\pm \sqrt{39i}\right]}{2}\right\}$

Answer 1

The correct option is A

$\left\{-6,1,\frac{\left[-5\pm \sqrt{39}i\right]}{2}\right\}$

Explanation for the correct option:

Find the solution set of the given equation

Given, $(x+1)(x+4)(x+2)(x+3)=120$

$$\begin{gathered} (x^2+4x+x+4)(x^2+3x+2x+6)=120 \\ (x^2+5x+4)(x^2+5x+6)=120 \end{gathered}$$

Let, $x^2+5x=m$

$$\begin{gathered} \Rightarrow (m+4)(m+6)=120 \\ \Rightarrow m^2+6m+4m+24=120 \\ \Rightarrow m^2+10m-96=0 \\ \Rightarrow m^2+16m-6m-96=0 \\ \Rightarrow m(m+16)-6(m+16)=0 \\ \Rightarrow \left(m-6\right)\left(m+16\right)=0 \\ (m-6)=0 \\ \Rightarrow m=6 \\ \left(m+16\right)=0 \\ \Rightarrow m=-16 \end{gathered}$$

Put the value of $m$back, we get equations

$x^2+5x-6=0\mathrm{and}{x}^2+5x+16=0$

Finding roots of $x^2+5\mathrm{x}-6=0$

$$\begin{gathered} {\mathrm{x}}^2+6\mathrm{x}-\mathrm{x}-6=0 \\ \Rightarrow \left(\mathrm{x}+6\right)\left(\mathrm{x}-1\right)=0 \\ x=-6\mathrm{and}x=1 \end{gathered}$$

Finding determinant of $x^2+5\mathrm{x}+16=0$

$$\begin{gathered} △=-39<0 \\ \Rightarrow \sqrt{△}=\sqrt{39i} \\ \Rightarrow x=\frac{\left[-5\pm \sqrt{39}i\right]}{2} \end{gathered}$$

Hence the correct option is option (A) i.e. $\left\{-6,1,\frac{\left[-5\pm \sqrt{39}i\right]}{2}\right\}$