The sum of all real roots of the equation $|x - 3{|^2} + |x - 3| - 2 = 0$
A. 2
B. 3
C. 4
D. 6
Last updated date: 29th Mar 2023
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Answer
307.8k+ views
Hint: : The first step is to simplify the given equation by taking the mod value and equating it to a variable, then to find the roots we will factorise the simplified equation and then we will find the roots.
Given equation is $|x - 3{|^2} + |x - 3| - 2 = 0$
To simplify the equation, let us consider $|x - 3| = a$,
${a^2} + a - 2 = 0$
Therefore, on factorising, we get,
$ \Rightarrow \left( {a + 2} \right)\left( {a - 1} \right) = 0$
Therefore,
$ \Rightarrow a = - 2\,or\,1,$
But, $|x - 3|$ cannot be negative number because we have a mod sign, therefore, we cannot consider $|x - 3|$= -2, therefore,
\[ \Rightarrow x = 4,2\]
On opening the mod, we get,
\[ \Rightarrow x - 3 = \pm 1\]
On solving, we get,
\[ \Rightarrow x = 4,2\]
Therefore, the sum of all the real roots is 4+2=6.
Option D is the correct answer.
Note: While solving these questions, try to simplify the given equation and then find the roots by factoring or using the formula of finding the roots whichever method is easier for that particular question.
Given equation is $|x - 3{|^2} + |x - 3| - 2 = 0$
To simplify the equation, let us consider $|x - 3| = a$,
${a^2} + a - 2 = 0$
Therefore, on factorising, we get,
$ \Rightarrow \left( {a + 2} \right)\left( {a - 1} \right) = 0$
Therefore,
$ \Rightarrow a = - 2\,or\,1,$
But, $|x - 3|$ cannot be negative number because we have a mod sign, therefore, we cannot consider $|x - 3|$= -2, therefore,
\[ \Rightarrow x = 4,2\]
On opening the mod, we get,
\[ \Rightarrow x - 3 = \pm 1\]
On solving, we get,
\[ \Rightarrow x = 4,2\]
Therefore, the sum of all the real roots is 4+2=6.
Option D is the correct answer.
Note: While solving these questions, try to simplify the given equation and then find the roots by factoring or using the formula of finding the roots whichever method is easier for that particular question.
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