Bluebook SAT Test Nonlinear Equations and Systems 5

Question 1

The line y = 2x + c is tangent to the parabola y = x² − 4x + 9. What is the value of c?

Accepted Answer

0

Answer

Setting the expressions equal gives x² − 4x + 9 = 2x + c, or x² − 6x + (9 − c) = 0. For the line to be tangent (exactly one intersection), the discriminant must equal zero: (−6)² − 4(1)(9 − c) = 0 → 36 − 36 + 4c = 0 → c = 0.

Question 2

How many real solutions does the system x² + y² = 4 and y = x² − 3 have?

Accepted Answer

4

Answer

Substituting y = x² − 3 into x² + y² = 4 gives x² + (x² − 3)² = 4, which simplifies to x⁴ − 5x² + 5 = 0. Let u = x²: the quadratic u² − 5u + 5 = 0 has discriminant 25 − 20 = 5 > 0, yielding two positive roots u = (5 ± √5)/2 ≈ 3.62 and 1.38. Each positive u gives two real x-values, so there are 4 real intersection points.

Question 3

The parabola y = x² + px + q and the line y = x + 5 intersect at two points whose x-coordinates r and s satisfy r + s = 4 and rs = −3. What is the value of p?

Accepted Answer

−3

Answer

Setting x² + px + q = x + 5 gives x² + (p − 1)x + (q − 5) = 0. By Vieta's formulas, the sum of the roots is r + s = −(p − 1). Setting −(p − 1) = 4 gives p − 1 = −4, so p = −3. (The product condition rs = q − 5 = −3 gives q = 2, but p is fully determined by the sum alone.)

Question 4

How many values of x satisfy the equation √(x + 5) = x − 1?

Accepted Answer

1

Answer

Squaring both sides gives x + 5 = (x − 1)² = x² − 2x + 1, so x² − 3x − 4 = 0, factoring as (x − 4)(x + 1) = 0 with solutions x = 4 and x = −1. Checking: x = 4 gives √9 = 3 = 4 − 1 ✓. But x = −1 gives √4 = 2 while x − 1 = −2, and a principal square root cannot equal a negative number, so x = −1 is extraneous. Only x = 4 is valid.

Question 5

For what integer value of k does the system y = x² − kx + k and y = x have exactly one solution?

Accepted Answer

1

Answer

Setting x² − kx + k = x gives x² − (k + 1)x + k = 0. The discriminant is (k + 1)² − 4k = k² + 2k + 1 − 4k = k² − 2k + 1 = (k − 1)². This is a perfect square and equals zero only when k = 1, which is the unique integer giving exactly one intersection point.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Nonlinear Equations and Systems 5