Question

The sum of roots of the equation \[{x^2} - \left| {2x - 3} \right| - 4 = 0\]is
A. \[2\]
B. \[ - 2\]
C. \[\sqrt 2 \]
D. \[ - \sqrt 2 \]

Answer 1

Hint: We take two cases for the equation as the term b is in modulus so we open up to two possibilities of the term being greater than or equal to zero and the term being less than zero.
Then we find four separate roots and discard the roots which contradict the condition of less than or greater than. In the end we calculate the sum of roots.

Complete step-by-step answer:
We have the equation \[{x^2} - \left| {2x - 3} \right| - 4 = 0\]
Now we know we can write \[\left| {2x - 3} \right| = 2x - 3\]when \[2x - 3 \geqslant 0\]and \[\left| {2x - 3} \right| = - (2x - 3)\]when \[2x - 3 < 0\]
CASE 1: When \[2x - 3 \geqslant 0\]
Solving for the value of x.
Add three to both sides of the inequality
\[ \Rightarrow 2x \geqslant 3\]
Divide both sides of inequality by 2
\[ \Rightarrow x \geqslant \dfrac{3}{2}\] … (1)
Now we solve the equation by substituting \[\left| {2x - 3} \right| = 2x - 3\]
\[
   \Rightarrow {x^2} - (2x - 3) - 4 = 0 \\
   \Rightarrow {x^2} - 2x + 3 - 4 = 0 \\
   \Rightarrow {x^2} - 2x - 1 = 0 \\
 \]
Comparing the equation to general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 1,b = - 2,c = - 1\].
Use the formula for finding the roots of a quadratic equation i.e. \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Substituting the values of \[a = 1,b = - 2,c = - 1\]in formula we get
\[
   \Rightarrow \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)( - 1)} }}{{2(1)}} \\
   \Rightarrow \dfrac{{2 \pm \sqrt {4 + 4} }}{2} \\
   \Rightarrow \dfrac{{2 \pm \sqrt 8 }}{2} \\
 \]
Put the value of \[\sqrt 8 = \sqrt {{2^2} \times 2} = 2\sqrt 2 \]
\[ \Rightarrow \dfrac{{2 \pm 2\sqrt 2 }}{2}\]
Take 2 common from the numerator
\[ \Rightarrow \dfrac{{2(1 \pm \sqrt 2 )}}{2}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow 1 \pm \sqrt 2 \]
So the two roots are \[1 + \sqrt 2 ,1 - \sqrt 2 \]
Substituting the value of \[\sqrt 2 = 1.414\]
Roots become \[2.414, - 0.414\]
From equation (1) \[x \geqslant \dfrac{3}{2}\]
So we reject the negative value
The root of the equation is \[1 + \sqrt 2 \]
CASE 2: When \[2x - 3 < 0\]
Solving for the value of x.
Add three to both sides of the inequality
\[ \Rightarrow 2x < 3\]
Divide both sides of inequality by 2
\[ \Rightarrow x < \dfrac{3}{2}\] … (2)
Now we solve the equation by substituting \[\left| {2x - 3} \right| = - (2x - 3)\]
\[
   \Rightarrow {x^2} - [ - (2x - 3)] - 4 = 0 \\
   \Rightarrow {x^2} + 2x - 3 - 4 = 0 \\
   \Rightarrow {x^2} + 2x - 7 = 0 \\
 \]
Comparing the equation to general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 1,b = 2,c = - 7\].
Use the formula for finding the roots of a quadratic equation i.e. \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Substituting the values of \[a = 1,b = 2,c = - 7\] in formula we get
\[
   \Rightarrow \dfrac{{ - (2) \pm \sqrt {{{(2)}^2} - 4(1)( - 7)} }}{{2(1)}} \\
   \Rightarrow \dfrac{{ - 2 \pm \sqrt {4 + 28} }}{2} \\
   \Rightarrow \dfrac{{ - 2 \pm \sqrt {32} }}{2} \\
 \]
Put the value of \[\sqrt {32} = \sqrt {{4^2} \times 2} = 4\sqrt 2 \]
\[ \Rightarrow \dfrac{{ - 2 \pm 4\sqrt 2 }}{2}\]
Take 2 common from the numerator
\[ \Rightarrow \dfrac{{2( - 1 \pm 2\sqrt 2 )}}{2}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow - 1 \pm 2\sqrt 2 \]
So the two roots are \[ - 1 + 2\sqrt 2 , - 1 - 2\sqrt 2 \]
Substituting the value of \[\sqrt 2 = 1.414\]
Roots become \[1.828, - 3.828\]
From equation (2) \[x < \dfrac{3}{2}\]
So we reject the value that is greater than \[\dfrac{3}{2}\]
The root of the equation is \[ - 1 - 2\sqrt 2 \].
Now we add the two roots
\[ - 1 - 2\sqrt 2 + 1 + \sqrt 2 = - 2\sqrt 2 + \sqrt 2 = - \sqrt 2 \]

So, option D is correct.

Note: Many students make mistake of opening up modulus as the positive value and write \[\left| {2x - 3} \right| = 2x - 3\] directly, which is wrong because the value of modulus depends on the fact that the value inside the modulus is negative or positive on the basis of which we should provide two cases.
Also, after we find the roots, we always have to check from the condition if the root is valid or not.