If k is the product of all real solutions to the equation |x^2 - 7| = 3x - 3, what is the … — GMAT

If $k$ is the product of all real solutions to the equation $|x^2 - 7| = 3x - 3$ , what is the value of $k$ ?

Answer: 8

The product of all real solutions is 8.

The equation

|x^2 - 7| = 3x - 3

requires that

3x - 3 \ge 0

, which gives the restriction

x \ge 1

. Solving the two cases

x^2 - 7 = 3x - 3

and

x^2 - 7 = -(3x - 3)

produces the candidate roots

x = 4, -1, 2, -5

. Eliminating the negative candidate roots leaves

x = 2

and

x = 4

as the only valid real solutions. Their product is

2 \times 4 = 8

Determine the domain constraint for valid solutions.

Since the absolute value expression

|x^2 - 7|

cannot be negative,

3x - 3 \ge 0

, which requires

x \ge 1

An absolute value quantity

|A|

is always non-negative, so any equation of the form

|A| = B

requires

B \ge 0

for real solutions.

Solve the positive case

x^2 - 7 = 3x - 3

and test for extraneous solutions.

x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0

, giving

x = 4

(valid, as

4 \ge 1

) and

x = -1

(extraneous, as

-1 < 1

Solutions must satisfy the non-negativity constraint of the right-hand side.

Solve the negative case

x^2 - 7 = -(3x - 3)

and test for extraneous solutions.

x^2 + 3x - 10 = 0 \implies (x + 5)(x - 2) = 0

, giving

x = 2

(valid, as

2 \ge 1

) and

x = -5

(extraneous, as

-5 < 1

Solutions must satisfy the non-negativity constraint of the right-hand side.

Compute the product of the valid real solutions.

k = 2 \times 4 = 8

The question asks for the product of all valid real solutions.

Absolute Value Equations with Variable Expressions and Extraneous Solution Elimination

Question