Question

Find the value of ‘\[x\]’, if given that: \[x + 3\left| x \right| = 10\].
A. \[\dfrac{5}{2}\]
B. \[ - 5\]
C. Does not exists
D. \[ - 5,\dfrac{5}{2}\]

Answer 1

Hint:Here, we will be going to consider the two values/roots for the variable ‘\[x\]’ say, ‘\[ - x\]’ and ‘\[ + x\]’ respectively in the given equation that is ‘\[x + 3\left| x \right| = 10\]’, as it implies the modulus sign in the given equation i.e. ‘\[\left| x \right|\]’. Hence, first of all, substituting the ‘\[x = - x\]’ in the given equation, then similarly, substituting ‘\[x = + x\]’ respectively and solving the respective equations mathematically, the desired value(s) is/are obtained.

Complete step by step answer:
Since, we have given that,
\[ \Rightarrow x + 3\left| x \right| = 10\] … (i)
As a result, ‘\[x\]’ has two possible roots that is in terms of ‘\[ - {\text{ve}}\]’ and ‘\[ + {\text{ve}}\]’ respectively, so as to find the value of ‘\[x\]’, … (\[\because \] due to the modulus sign)
Hence, first of all, considering root of ‘\[x\]’ equals to ‘\[ - x\]’ in equation (i), we get
\[ \Rightarrow x + 3\left( { - x} \right) = 10\]
Solving the equation mathematically, we get
\[ \Rightarrow x - 3x = 10\]
\[ \Rightarrow - 2x = 10\]
Dividing the above equation by ‘\[2\]’, we get
\[ \Rightarrow x = \dfrac{{10}}{{ - 2}}\]
Simplifying the equation, we get
\[ \Rightarrow x = - 5\] … (ii)
Similarly,

Considering the root of ‘\[x\]’ equals to ‘\[ + x\]’ in equation (i), we get
\[ \Rightarrow x + 3\left( { + x} \right) = 10\]
Solving the equation mathematically, we get
\[ \Rightarrow x + 3x = 10\]
Adding the above equation, we get
\[ \Rightarrow 4x = 10\]
\[ \Rightarrow x = \dfrac{{10}}{4}\]
Simplifying the value in simplest form (by dividing by its common divisible that is \[2\]), we get
\[ \Rightarrow x = \dfrac{5}{2}\] … (iii)
From (ii) and (iii), the value of ‘\[x\]’ seems to be the two possible values that is ‘\[ - 5\]’ or ‘\[\dfrac{5}{2}\]’ respectively.

Hence, option D is correct.

Note:One must able to remember the definition of modulus while solving any algebraic solution or geometric terms that the presence of modulus (must) exists two possible cases/roots in terms of ‘positive’ and ‘negative’ respectively, so as to be sure of our final answer.