Determine the sum of all real solutions to the equation |x^2 - 4x| = 3x - 6. — GMAT

Determine the sum of all real solutions to the equation $|x^2 - 4x| = 3x - 6$ .

The sum of all valid real solutions is

9

The correct answer is

9

. Setting up the two cases

x^2 - 4x = 3x - 6

and

x^2 - 4x = -(3x - 6)

yields candidate roots

x = 1, 6, 3,

and

-2

. Because the absolute value expression

|x^2 - 4x|

cannot be negative,

3x - 6

must be non-negative, requiring

x \ge 2

. Evaluating each candidate shows that

x = 1

and

x = -2

produce negative right-hand sides and are extraneous. The only valid solutions are

x = 3

and

x = 6

, whose sum is

3 + 6 = 9

Establish the domain condition for the right-hand side of the absolute value equation.

Since absolute values are non-negative,

|x^2 - 4x| \geq 0

requires

3x - 6 \geq 0 \implies x \geq 2

An absolute value expression cannot equal a negative number.

Solve Case 1 where

x^2 - 4x = 3x - 6

x^2 - 7x + 6 = 0 \implies (x - 1)(x - 6) = 0 \implies x = 1

x = 6

This corresponds to the positive branch of the absolute value.

Solve Case 2 where

x^2 - 4x = -(3x - 6)

x^2 - 4x = -3x + 6 \implies x^2 - x - 6 = 0 \implies (x - 3)(x + 2) = 0 \implies x = 3

x = -2

This corresponds to the negative branch of the absolute value.

Test all candidate solutions (

x = -2, 1, 3, 6

) against the domain constraint

x \geq 2

x = -2

yields

3(-2)-6 = -12 < 0

(extraneous).

x = 1

yields

3(1)-6 = -3 < 0

(extraneous).

x = 3

yields

|9-12| = 3 = 3(3)-6

(valid).

x = 6

yields

|36-24| = 12 = 3(6)-6

(valid).

Extraneous roots introduced by unconstrained case splitting must be eliminated.

Sum the valid real solutions.

3 + 6 = 9

The question asks specifically for the sum of all valid real solutions.

Absolute Value Equations and Extraneous Solution Verification

Question