PSAT Math Quiz: Systems Of Polynomial Equations

What this quiz covers

This quiz focuses on Systems Of Polynomial Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $0$ (correct answer)
2. $2$
3. $4$
4. $8$

Explanation: We need the product of x-coordinates where the line y = ½x + 1 meets the parabola y = -½x² + 2x + 1. This system combines a line and parabola, solved by substitution. Setting ½x + 1 = -½x² + 2x + 1, we get ½x + 1 + ½x² - 2x - 1 = 0, which simplifies to ½x² - (3/2)x = 0, or ½x(x - 3) = 0. This gives x = 0 or x = 3, so the product of x-coordinates is 0 × 3 = 0. A key insight: when one root is zero, the product of all roots must be zero. For any quadratic ax² + bx + c = 0, Vieta's formula gives the product of roots as c/a, and here c = 0 confirms our answer.

Question 2

1. 0
2. 1
3. 2 (correct answer)
4. 3

Explanation: The question asks for the number of solutions to the system of y = x + 2 and y = x², which represents the intersection points of a line and a parabola on the coordinate plane. This system includes one linear and one nonlinear equation, solved by substitution: set x² = x + 2, rearrange to x² - x - 2 = 0, and factor as (x - 2)(x + 1) = 0, giving x = 2 and x = -1. These yield two distinct real solutions, confirming two intersection points. A key error could be incorrectly factoring the quadratic or overlooking the discriminant (here, 1 + 8 = 9 > 0, indicating two real roots). Note that a parabola and a line can intersect 0, 1, or 2 times, and in this case, it's two; a useful strategy is to calculate the discriminant of the resulting quadratic to quickly determine the number of solutions without fully solving.

Question 3

1. 0
2. 1
3. 2 (correct answer)
4. 4

Explanation: The question asks how many solutions the system has, meaning how many times the drone's height model intersects the safety limit y=0. This involves setting a quadratic equal to a constant (horizontal line), solvable by finding roots of the quadratic equation, with 0, 1, or 2 possible intersections. Set $x^2 - 6x + 8 = 0$. Factor as $(x - 2)(x - 4) = 0$, giving $x = 2$ and $x = 4$, so two solutions. A common error is misfactoring or ignoring the constant term. A test-taking strategy is to compute the discriminant $(36 - 32 = 4 > 0)$ to confirm two real roots quickly.

Question 4

1. $(1,3)$ only
2. $(0,1)$ and $(4,9)$ (correct answer)
3. $(0,1)$ only
4. $(1,3)$ and $(5,11)$

Explanation: The question asks for the points where two models of a ball's height agree, meaning the intersection points of the quadratic equation y = -x² + 6x + 1 and the linear equation y = 2x + 1. This is a system consisting of one linear and one nonlinear (quadratic) equation, which can be solved algebraically by setting the equations equal to each other. Set -x² + 6x + 1 = 2x + 1, subtract 2x + 1 from both sides to get -x² + 4x = 0, then multiply by -1 to obtain x² - 4x = 0, and factor as x(x - 4) = 0, yielding x = 0 or x = 4. Substitute these x-values back into y = 2x + 1 to find y = 1 when x = 0 and y = 9 when x = 4, so the points are (0, 1) and (4, 9). A common error is mishandling the signs during subtraction, which might lead to an incorrect quadratic equation. Remember that a line and a parabola can intersect at 0, 1, or 2 points, and here they intersect at two; for test-taking, verify solutions by plugging back into both original equations.

Question 5

1. $(1,3)$ only
2. $(0,1)$ and $(4,9)$ (correct answer)
3. $(0,1)$ and $(2,5)$
4. $(1,3)$ and $(3,7)$

Explanation: We need to find where the ball's height h = -t² + 6t + 1 equals the sensor's reading h = 2t + 1. This is a system with a parabola (quadratic) and a line, which we solve by substitution. Setting -t² + 6t + 1 = 2t + 1, we get -t² + 6t + 1 - 2t - 1 = 0, which simplifies to -t² + 4t = 0, or -t(t - 4) = 0. This gives t = 0 or t = 4. When t = 0, h = 2(0) + 1 = 1, and when t = 4, h = 2(4) + 1 = 9, so our solutions are (0,1) and (4,9). A common error is forgetting to find the h-values after solving for t. Since we found two distinct intersection points, this confirms that a parabola and line can intersect at 0, 1, or 2 points.

Question 6

1. $-4$
2. $0$
3. $4$ (correct answer)
4. $8$

Explanation: We need to find the product of x-coordinates where $y = x^2 - 6x + 8$ and $y = -2x + 4$ intersect. This quadratic-linear system uses substitution. Setting equal: $x^2 - 6x + 8 = -2x + 4$, which becomes $x^2 - 4x + 4 = 0$. This factors as $(x - 2)^2 = 0$, giving $x = 2$ as a repeated root. With only one x-value (multiplicity 2), we have $x_1 = x_2 = 2$, so $x_1 x_2 = 2 imes 2 = 4$. A common error is thinking one solution means the product is undefined, but a repeated root counts twice. For any quadratic $ax^2 + bx + c = 0$, the product of roots equals $c/a$, which here is $4/1 = 4$.

Question 7

1. $-2$
2. $0$
3. $2$ (correct answer)
4. $4$

Explanation: We need to find where $y = x^2 - 4$ and $y = 2x$ intersect, then sum the x-coordinates. This quadratic-linear system is solved by substitution. Setting equal: $x^2 - 4 = 2x$, which becomes $x^2 - 2x - 4 = 0$. Using the quadratic formula: $x = (2 \pm \sqrt{4 + 16})/2 = (2 \pm \sqrt{20})/2 = (2 \pm 2\sqrt{5})/2 = 1 \pm \sqrt{5}$. The sum of the x-coordinates is $(1 + \sqrt{5}) + (1 - \sqrt{5}) = 2$. A useful property: for any quadratic $ax^2 + bx + c = 0$, the sum of roots equals $-b/a$, which here is $-(-2)/1 = 2$. This provides a quick check without finding individual roots.

Question 8

1. $(1,0)$ and $(4,3)$ (correct answer)
2. $(0,-1)$ and $(3,2)$
3. $(1,0)$ only
4. $(2,1)$ and $(3,2)$

Explanation: The question seeks the intersection points of the quadratic ramp model and linear support beam. This system combines a linear and a nonlinear quadratic equation, solved via substitution, with the reminder that such graphs can intersect at 0, 1, or 2 points. Set x² - 4x + 3 = x - 1, simplify to x² - 5x + 4 = 0, and factor as (x - 1)(x - 4) = 0, so x = 1 and x = 4. Corresponding y-values are y = 0 for x = 1 and y = 3 for x = 4, giving (1, 0) and (4, 3). Errors often occur in algebraic simplification, like incorrect subtraction. When time is limited, graph mentally or check discriminants to confirm two real roots.

Question 9

1. $(1,3)$ only
2. $(0,1)$ and $(4,9)$ (correct answer)
3. $(0,1)$ only
4. $(2,5)$ and $(3,7)$

Explanation: The question asks for all solutions (t, h) where the quadratic model for a ball's height agrees with the linear sensor calibration. This is a system of one linear and one nonlinear (quadratic) equation, which we solve by substitution, noting that a line and a parabola can intersect at 0, 1, or 2 points. Set -t² + 6t + 1 = 2t + 1, simplify to -t² + 4t = 0, and factor as -t(t - 4) = 0, yielding t = 0 and t = 4. Then, substitute into the linear equation: for t = 0, h = 1; for t = 4, h = 9, giving points (0, 1) and (4, 9). A common error is forgetting to check both solutions in the original equations, but both satisfy here. For test-taking, verify solutions by plugging back in to avoid calculation mistakes.

Question 10

1. 0
2. 1
3. 2 (correct answer)
4. 3

Explanation: The question asks for the number of solutions to the system, which is the number of intersection points between the line and the parabola. This system involves one linear and one quadratic equation, solved by setting them equal, remembering a parabola and line can intersect 0, 1, or 2 times. Set $x + 1 = x^2 - 2x - 3$, rearrange to $x^2 - 3x - 4 = 0$. Factor as $(x - 4)(x + 1) = 0$, yielding $x = 4$ and $x = -1$, so two solutions. A key error could be incorrect rearrangement of terms, resulting in wrong coefficients. A test-taking strategy is to calculate the discriminant ($9 + 16 = 25 > 0$) to quickly confirm there are two real roots without fully solving.

Question 11

1. $-1$
2. $0$ (correct answer)
3. $1$
4. $3$

Explanation: The question asks for the product of the $x$-coordinates of all solutions to the system of concentration models. This is a linear and quadratic system, solved by substitution, with possible 0, 1, or 2 intersections between line and parabola. Set $x^2 + 2x - 3 = 3x - 3$, rearrange to $x^2 - x = 0$. Factor as $x(x - 1) = 0$, giving $x = 0$ and $x = 1$, with product $0 \times 1 = 0$. A common error is failing to move all terms to one side, resulting in an incorrect equation. For efficiency, note that for $ax^2 + bx + c = 0$, the product of roots is $c/a = 0/1 = 0$, confirming the answer without finding individual roots.

Question 12

1. 0 solutions
2. 1 solution (correct answer)
3. 2 solutions
4. Infinitely many

Explanation: The question asks about the number of solutions to the system, where the student claims the line touches the parabola at one point. This is setting a quadratic equal to zero (horizontal line), solved by finding roots, with a parabola intersecting a line 0, 1, or 2 times, but here possibly a tangent case. Set $x^2 - 4x + 4 = 0$. Factor as $(x - 2)^2 = 0$, giving a double root at $x = 2$, which counts as one solution. A key point is recognizing a perfect square indicates tangency and one intersection. A strategy is to check the discriminant ($16 - 16 = 0$), confirming exactly one real solution.

Question 13

1. $(0,-1)$ and $(1,0)$
2. $(-2,-3)$ and $(1,0)$ (correct answer)
3. $(-1,-2)$ and $(2,1)$
4. $(-2,-3)$ and $(0,-1)$

Explanation: The question asks for all solutions to the system where the park path parabola meets the straight fence line. This system consists of one linear and one quadratic equation, solved by substitution, recalling a line and parabola can intersect at 0, 1, or 2 points. Set $-x^2 + 1 = x - 1$, rearrange to $-x^2 - x + 2 = 0$, then multiply by -1 to get $x^2 + x - 2 = 0$. Factor as $(x + 2)(x - 1) = 0$, giving $x = -2$ and $x = 1$; substitute to find $y = -3$ and $y = 0$, yielding $(-2, -3)$ and $(1, 0)$. A key error is mishandling signs during rearrangement. A test-taking strategy is to verify by plugging solutions back into both equations and noting the discriminant ($1 + 8 = 9 > 0$) indicates two real intersections.

Question 14

1. 0
2. 1
3. 2 (correct answer)
4. 3

Explanation: We need to find how many times the parabola $y = x^2 - 4x - 5$ intersects the line $y = -x - 1$. This system combines a quadratic with a linear equation, solved by substitution. Setting equal: $x^2 - 4x - 5 = -x - 1$, which becomes $x^2 - 3x - 4 = 0$. Using the quadratic formula or factoring: $(x - 4)(x + 1) = 0$, giving $x = 4$ and $x = -1$. Since we found two distinct x-values, there are 2 intersection points. A key insight is that a parabola and line can intersect 0, 1, or 2 times—finding two solutions means we've found all possible intersections. When counting solutions, remember each x-value corresponds to exactly one point $(x, y)$.

Question 15

1. $-3$ (correct answer)
2. $-1$
3. $1$
4. $3$

Explanation: We need to find the sum of x-coordinates where $y = x^2 + 2x - 3$ and $y = 1 - x$ intersect. This quadratic-linear system uses substitution. Setting equal: $x^2 + 2x - 3 = 1 - x$, which becomes $x^2 + 3x - 4 = 0$. Factoring: $(x + 4)(x - 1) = 0$, giving $x = -4$ or $x = 1$. The sum of x-coordinates is -4 + 1 = -3. For verification, the sum of roots of $ax^2 + bx + c = 0$ equals $-b/a$, which here is $-3/1 = -3$. When dealing with sums of roots, this formula provides a quick alternative to finding individual solutions.

Question 16

1. $(0,1)$ and $(4,9)$ (correct answer)
2. $(1,3)$ only
3. $(0,1)$ only
4. $(2,5)$ and $(3,7)$

Explanation: This problem asks for all points where the parabola $y = -x^2 + 6x + 1$ and the line $y = 2x + 1$ intersect. This is a system with one quadratic and one linear equation, which we solve by substitution. Setting the equations equal: $-x^2 + 6x + 1 = 2x + 1$, which simplifies to $-x^2 + 4x = 0$, or $-x(x - 4) = 0$. This gives $x = 0$ or $x = 4$. When $x = 0$, $y = 2(0) + 1 = 1$, and when $x = 4$, $y = 2(4) + 1 = 9$. A common error is forgetting to find the y-values after solving for x, but the question asks for complete coordinate pairs. Since a parabola and line can intersect at most twice, finding two solutions confirms we have all of them.

Question 17

1. $2$
2. $4$
3. $6$
4. $8$ (correct answer)

Explanation: The question asks for all solutions to the system y = -x² + 8x - 12 and y = 0, which are the x-intercepts of the parabola, and then the sum of those x-values. This is a system of a quadratic (nonlinear) equation and the linear equation y = 0, solved by setting -x² + 8x - 12 = 0, multiplying by -1 to get x² - 8x + 12 = 0, and factoring as (x - 2)(x - 6) = 0, so x = 2 and x = 6 with corresponding points (2, 0) and (6, 0). The sum of the x-values is 2 + 6 = 8. A common mistake is forgetting to multiply by -1, leading to incorrect roots or sum. Remember, a horizontal line like y = 0 can intersect a parabola at 0, 1, or 2 points; for efficiency in testing, use the sum of roots formula (-b/a = 8) directly from the quadratic x² - 8x + 12 = 0.

Question 18

1. $-4$
2. $-1$
3. $1$ (correct answer)
4. $4$

Explanation: The question asks for the sum of the x-coordinates of the intersection points of y = x + 4 and y = x², assuming both solutions exist. This system of linear and nonlinear equations is solved by setting x² = x + 4, rearranging to x² - x - 4 = 0. The sum of the roots is -b/a = 1, from the quadratic coefficients, without needing to find individual roots (discriminant 1 + 16 = 17 > 0 confirms two real solutions). A key error might be misidentifying the coefficients for the sum formula. Note that a parabola and line intersect at most twice, which they do here; a test-taking tip is to use Vieta's formulas for sum and product of roots to answer quickly without solving fully.

Question 19

1. $(0,-9)$ and $(3,0)$ (correct answer)
2. $(0,-9)$ only
3. $(3,0)$ only
4. $(1,-6)$ and $(9,18)$

Explanation: We need all points where y = $x^2 - 9$ and y = $3x - 9$ intersect. This is a quadratic-linear system solved by substitution. Setting equal: $x^2 - 9 = 3x - 9$, which simplifies to $x^2 = 3x$, or $x^2 - 3x = 0$. Factoring: $x(x - 3) = 0$, giving $x = 0$ or $x = 3$. When $x = 0$: y = $3(0) - 9$ = -9, and when $x = 3$: y = $3(3) - 9$ = 0. The solutions are (0, -9) and (3, 0). A common error is solving $x^2 = 3x$ by dividing both sides by x, which loses the x = 0 solution—always factor instead to preserve all solutions.

Question 20

1. $(1,3)$ and $(4,6)$ (correct answer)
2. $(2,4)$ only
3. $(1,3)$ only
4. $(2,4)$ and $(3,5)$

Explanation: The question seeks all solutions (x, P) where two profit models agree. This system of linear and nonlinear quadratic equations is solved by substitution, with potential for 0, 1, or 2 intersections between line and parabola. Set x + 2 = x² - 4x + 6, rearrange to x² - 5x + 4 = 0, factor as (x - 1)(x - 4) = 0, giving x = 1 and x = 4. Then P = 3 and P = 6, so (1, 3) and (4, 6). A key error is failing to substitute back for P-values. Verify by plugging into originals to ensure accuracy.