Bluebook SAT Test Nonlinear Equations and Systems 2

Question 1

Which values of x satisfy x² − x − 12 = 0?

Accepted Answer

x = 4 and x = −3

Answer

Factor: (x−4)(x+3) = 0 → x = 4 or x = −3.

Question 2

For the equation 3x² − 12x + 12 = 0, what is the value of the discriminant?

Accepted Answer

0

Answer

Discriminant = (−12)² − 4(3)(12) = 144 − 144 = 0.

Question 3

Using the quadratic formula, what are the solutions of x² − 6x + 7 = 0?

Accepted Answer

x = 3 ± √2

Answer

x = [6 ± √(36−28)] / 2 = [6 ± √8] / 2 = [6 ± 2√2] / 2 = 3 ± √2.

Question 4

The system y = x² and y = x + 6 is solved by substitution. What are the x-values of the intersection points?

Accepted Answer

x = 3 and x = −2

Answer

Set x² = x + 6 → x² − x − 6 = 0 → (x−3)(x+2) = 0 → x = 3 or x = −2.

Question 5

How many points of intersection does the system y = x² − 3 and y = 2x have?

Accepted Answer

2

Answer

x² − 3 = 2x → x² − 2x − 3 = 0; discriminant = 4 + 12 = 16 > 0, so 2 intersections.

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