Question

If \[|{{x}^{2}}-x-6|=x+2\], then the values of x are
(a) -2, 2, -4
(b) -2, 2, 4
(c) 3, 2, -2
(d) 4, 4, 3

Answer 1

Hint: We will take case 1 first where the left hand side of the given equation \[|{{x}^{2}}-x-6|=x+2\] is less than 0 and solving this we get a domain. Then we solve the whole expression and see if any of the values belongs to the domain or not. If any of the values belongs, those values are our answer. Similarly we will take case 2 where the left hand side of the expression is greater than equal to 0 and will similarly solve the rest as in case 1.

Complete step-by-step answer:
So we will first take case 1 and assume that the expression inside modulus is less than zero. Using this information, we get,
Case 1: \[{{x}^{2}}-x-6<0.......(1)\]
So now factoring in equation (1) we get,
\[\Rightarrow {{x}^{2}}-3x+2x-6<0.......(2)\]
Now taking common terms out in equation (2) we get,
\[\begin{align}
  & \Rightarrow x(x-3)+2(x-3)<0 \\
 & \Rightarrow (x+2)(x-3)<0........(3) \\
\end{align}\]
So from equation (3) we can say that \[-2 < x < 3.......(4)\].
In this case the equation given in the question becomes,
\[\Rightarrow -({{x}^{2}}-x-6)=x+2........(5)\]
Now multiply both sides of equation (5) by minus we get,
\[\Rightarrow {{x}^{2}}-x-6=-x-2........(6)\]
Now cancelling similar terms from equation (6) and rearranging we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}-6=-2 \\
 & \Rightarrow {{x}^{2}}=6-2 \\
 & \Rightarrow {{x}^{2}}=4 \\
 & \Rightarrow x=-2,2.........(7) \\
\end{align}\]
Now from equation (7) only 2 belongs to the domain in equation (4). Hence 2 is the solution in this case.
So we will first take case 2 and assume that the expression inside modulus is greater than or equal to zero. Using this information, we get,
Case 1: \[{{x}^{2}}-x-6\ge 0.......(8)\]
So now factoring in equation (8) we get,
\[\Rightarrow {{x}^{2}}-3x+2x-6\ge 0.......(9)\]
Now taking common terms out in equation (9) we get,
\[\begin{align}
  & \Rightarrow x(x-3)+2(x-3)\ge 0 \\
 & \Rightarrow (x+2)(x-3)\ge 0........(10) \\
\end{align}\]
So from equation (10) we can say that \[x\le -2,\,x\ge 3.......(11)\].
In this case the equation given in the question becomes,
\[\Rightarrow {{x}^{2}}-x-6=x+2........(12)\]
Now rearranging both sides of equation (12) we get,
\[\Rightarrow {{x}^{2}}-2x-8=0........(13)\]
Now factoring in equation (13) and rearranging we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}-4x+2x-8=0 \\
 & \Rightarrow x(x-4)+2(x-4)=0 \\
 & \Rightarrow (x+2)(x-4)=0 \\
 & \Rightarrow x=-2,4.........(14) \\
\end{align}\]
Now from equation (14) both these values belong to the domain in equation (11). Hence -2 and 4 is the solution in this case.
Hence the final solution for x is 2, -2, 4. Thus the correct answer is option (b).

Note: Remembering properties of modulus is the key here. |x| is equal to –x when x is negative and |x| is equal to x when x is positive. Checking the values if it lies in the domain is important. Also in a hurry we can make a mistake in solving equation (6) and equation (12) and hence we need to be careful while solving these equations.