Question

The number of elements in the set $x\in R:(\left|x\right|-3)\left|x+4\right|=6$

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Step 1: Rewrite the given equation.

An equation $(\left|x\right|-3)\left|x+4\right|=6$ is given.

Rewrite the equation as follows:

When $x\in (-\infty ,-4]$ the given equation becomes.

$$\begin{gathered} (-x-3)\left(-x-4\right)=6 \\ \Rightarrow (x+3)\left(x+4\right)=6 \\ \Rightarrow x^2+7x+12=6 \\ \Rightarrow x^2+7x+6=0 \end{gathered}$$

Therefore, the given equation becomes $x^2+7x+6=0$ when $x\in (-\infty ,-4]$.

When $x\in (-4,0)$ the given equation becomes.

$$\begin{gathered} (-x-3)\left(x+4\right)=6 \\ \Rightarrow -(x+3)\left(x+4\right)=6 \\ \Rightarrow x^2+7x+12=-6 \\ \Rightarrow x^2+7x+18=0 \end{gathered}$$

Therefore, the given equation becomes $x^2+7x+18=0$ when $x\in (-4,0)$.

When $x\in [0,\infty )$ the given equation becomes.

$$\begin{gathered} (x-3)\left(x+4\right)=6 \\ \Rightarrow x^2+x-12=6 \\ \Rightarrow x^2+x-18=0 \end{gathered}$$

Therefore, the given equation becomes $x^2+x-18=0$ when $x\in [0,\infty )$.

Step 2: Find the number of solutions of the given equation.

When $x\in (-\infty ,-4]$ is given equation is $x^2+7x+6=0$.

Use quadratic formula.

$$\begin{gathered} x=\frac{-7\pm \sqrt{7^2-4\left(6\right)}}{2} \\ \Rightarrow x=\frac{-7\pm \sqrt{49-24}}{2} \\ \Rightarrow x=\frac{-7\pm \sqrt{25}}{2} \\ \Rightarrow x=\frac{-7\pm 5}{2} \\ \Rightarrow x=-6,-1 \end{gathered}$$

But $x\in (-\infty ,-4]$. Therefore, $x=-6$ is the only solution when $x$ is less than or equal to $-4$.

When $x\in (-4,0)$ is given equation is $x^2+7x+18=0$.

Find the discriminant of the above quadratic equation as follows:

$$\begin{gathered} D=7^2-4\left(18\right) \\ \Rightarrow D=7^2-4\left(18\right)<0 \end{gathered}$$

Therefore, there is no solution when $x\in (-4,0)$.

When $x\in [0,\infty )$ is given equation is $x^2+x-18=0$.

Use quadratic formula.

$$\begin{gathered} x=\frac{-1\pm \sqrt{1^2-4(-18)}}{2} \\ \Rightarrow x=\frac{-1\pm \sqrt{1+72}}{2} \\ \Rightarrow x=\frac{-7\pm \sqrt{73}}{2} \\ \Rightarrow x=\frac{-7-\sqrt{73}}{2},\frac{-7+\sqrt{73}}{2} \end{gathered}$$

But $x\in [0,\infty )$. Therefore, $\Rightarrow x=\frac{-7+\sqrt{73}}{2}$ is the only solution when $x$ is greater than or equal to $0$.

Since, the total number of solutions for the given equation is $2$.

Therefore, there are two elements in the given set.

Hence, option $A$ is the correct answer.