SAT 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 SAT 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. $-3$
2. $9$
3. $0$
4. $3$ (correct answer)

Explanation: We solve $y = x^2 - 9$ and $y = 3x$ to find intersection points, then sum the x-coordinates. Setting $x^2 - 9 = 3x$ gives $x^2 - 3x - 9 = 0$. For any quadratic $ax^2 + bx + c = 0$, the sum of roots equals $-b/a$. Here, a = 1 and b = -3, so the sum is $-(-3)/1 = 3$. We can verify by solving: using the quadratic formula gives $x = \frac{3 \pm \sqrt{9+36}}{2} = \frac{3 \pm \sqrt{45}}{2} = \frac{3 \pm 3\sqrt{5}}{2}$. The sum is $\frac{3 + 3\sqrt{5}}{2} + \frac{3 - 3\sqrt{5}}{2} = \frac{6}{2} = 3$. This problem specifically tests knowledge of the sum of roots formula, avoiding the need to find individual solutions.

Question 2

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

Explanation: This problem asks for solutions to $y = x^2 + 2x - 8$ and $y = 0$, which means finding where the parabola crosses the x-axis. Setting $x^2 + 2x - 8 = 0$, we can factor by finding two numbers that multiply to -8 and add to 2: these are 4 and -2. So $(x + 4)(x - 2) = 0$, giving $x = -4$ and $x = 2$. The solutions are $(-4, 0)$ and $(2, 0)$, representing the x-intercepts of the parabola. A common error is to confuse the signs when factoring; remember that if $(x + 4) = 0$, then $x = -4$. When $y = 0$, we're specifically looking for x-intercepts, which is a special case of a parabola-line system.

Question 3

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

Explanation: The question asks for the number of solutions, or intersection points, between a parabolic fountain stream and a linear walkway edge. Solve this quadratic-linear system by substitution, noting a parabola and line can intersect 0, 1, or 2 times. Set $-x^2 + 4x = x$, simplify to $-x^2 + 3x = 0$ or $x(3 - x) = 0$, yielding $x = 0$ and $x = 3$. Both give real y-values, confirming two intersections. Errors include discarding x = 0 as non-physical without checking. Quickly factoring quadratics can help count solutions efficiently on exams.

Question 4

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

Explanation: We need to find where the parabola meets the horizontal threshold line, then find the product of x-coordinates. The system has y = $x^2 + 2x + 1$ and y = 3, so we solve $x^2 + 2x + 1 = 3$, which simplifies to $x^2 + 2x - 2 = 0$. For a quadratic $ax^2 + bx + c = 0$, Vieta's formulas tell us that the product of roots $x_1 x_2 = c/a = -2/1 = -2$. This is a time-saving technique—we don't need to find individual x-values when only their product is requested. A common error is solving for individual roots when Vieta's formulas provide a direct answer. For any quadratic equation, the product of roots equals c/a.

Question 5

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

Explanation: We need to find where the ball's parabolic path intersects with the drone's linear path. This is a system with one quadratic equation (parabola) and one linear equation, which can be solved by substitution. Since both equations equal y, we set them equal: $-x^2 + 6x - 5 = x + 1$, which simplifies to $-x^2 + 5x - 6 = 0$ or $x^2 - 5x + 6 = 0$. Factoring gives $(x - 1)(x - 4) = 0$, so $x = 1$ or $x = 4$. When $x = 1$, $y = 1 + 1 = 2$; when $x = 4$, $y = 4 + 1 = 5$. A common error is to stop after finding x-values without calculating the corresponding y-values. Remember that a parabola and line can intersect at 0, 1, or 2 points—here we have 2 intersection points.

Question 6

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

Explanation: This asks for the number of solutions when a parabola meets a line. We have a quadratic equation $y = x^2 - 4x + 3$ and a linear equation $y = 2x - 1$, so we use substitution. Setting them equal: $x^2 - 4x + 3 = 2x - 1$, which simplifies to $x^2 - 6x + 4 = 0$. Using the discriminant $b^2 - 4ac = (-6)^2 - 4(1)(4) = 36 - 16 = 20$, which is positive, indicating 2 real solutions. A common mistake is assuming all parabola-line systems have 2 solutions, but the discriminant determines whether there are 0, 1, or 2 intersection points. When the discriminant is positive (as here), there are exactly 2 solutions.

Question 7

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

Explanation: We need to find how many times the projectile's parabolic path reaches the horizontal sensor beam height. This system has a quadratic $y = -x^2 + 2x + 3$ and a horizontal line $y = 4$, solvable by substitution. Setting them equal: $-x^2 + 2x + 3 = 4$, which simplifies to $-x^2 + 2x - 1 = 0$ or $x^2 - 2x + 1 = 0$. This factors as $(x - 1)^2 = 0$, giving $x = 1$ as a repeated root, meaning the parabola just touches the line at one point. A common error is assuming that because a quadratic equation is involved, there must be 2 solutions—but when the discriminant equals 0, there's only 1 solution (tangent point). Remember: parabola and line can intersect at 0, 1, or 2 points.

Question 8

1. $(0,2)$ and $(2,3)$
2. $(0,2)$ and $(\tfrac{5}{2},\tfrac{13}{4})$ (correct answer)
3. $(-\tfrac{1}{2},\tfrac{7}{4})$ and $(2,3)$
4. $(0,2)$ only

Explanation: We need to find where the flashlight beam line intersects the curved mirror edge. The system has $y = \frac{1}{2}x + 2$ and $y = x^2 - 2x + 2$, solvable by substitution. Setting them equal: $\frac{1}{2}x + 2 = x^2 - 2x + 2$, which simplifies to $\frac{1}{2}x = x^2 - 2x$, then $0 = x^2 - 2x - \frac{1}{2}x = x^2 - \frac{5}{2}x$. Factoring: $x(x - \frac{5}{2}) = 0$, so $x = 0$ or $x = \frac{5}{2}$. When $x = 0$, $y = \frac{1}{2}(0) + 2 = 2$; when $x = \frac{5}{2}$, $y = \frac{1}{2}(\frac{5}{2}) + 2 = \frac{5}{4} + 2 = \frac{13}{4}$. The solutions are $(0, 2)$ and $(\frac{5}{2}, \frac{13}{4})$. A common error with fractions is arithmetic mistakes—always double-check calculations involving fractions on the SAT.

Question 9

1. $(2,0)$ and $(4,8)$ (correct answer)
2. $(0,-8)$ and $(6,16)$
3. $(2,0)$ only
4. $(4,8)$ only

Explanation: We need to find where the car's stopping distance parabola intersects the safety guideline line. This system has a quadratic equation $y = x^2 - 2x$ and a linear equation $y = 4x - 8$, solvable by substitution. Setting them equal: $x^2 - 2x = 4x - 8$, which simplifies to $x^2 - 6x + 8 = 0$. Factoring gives $(x - 2)(x - 4) = 0$, so $x = 2$ or $x = 4$. When $x = 2$, $y = 4(2) - 8 = 0$; when $x = 4$, $y = 4(4) - 8 = 8$, giving us points $(2, 0)$ and $(4, 8)$. A common error is mixing up the x and y coordinates or only finding one solution when there are two. Always verify your solutions by substituting back into both original equations.

Question 10

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

Explanation: This problem asks for solutions to $y = (x - 2)^2$ and $y = 4$, a system with a parabola and horizontal line. Since $y = 4$ in both equations, we solve $(x - 2)^2 = 4$ by taking square roots of both sides. This gives $x - 2 = \pm 2$, so $x - 2 = 2$ or $x - 2 = -2$, yielding $x = 4$ or $x = 0$. The solutions are $(0, 4)$ and $(4, 4)$, representing where the parabola reaches height 4. A key insight is that a parabola intersects a horizontal line at two points when the line cuts through the parabola above its vertex, at one point if it touches the vertex, or at zero points if below. Always remember to consider both positive and negative square roots.

Question 11

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

Explanation: We need to find when the ball and drone are at the same height, which occurs where the parabola y = -x² + 6x + 1 intersects the line y = 2x + 1. This is a system with one quadratic and one linear equation, so we'll use substitution. Setting the equations equal: -x² + 6x + 1 = 2x + 1, which simplifies to -x² + 4x = 0, then -x(x - 4) = 0, giving x = 0 or x = 4. When x = 0, y = 2(0) + 1 = 1; 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 intersection points.

Question 12

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

Explanation: To find where the parabola y = -x² + 4x meets the line y = x, we solve this system using substitution. Setting equal: -x² + 4x = x, which simplifies to -x² + 3x = 0, then -x(x - 3) = 0. This gives x = 0 or x = 3. When x = 0, y = 0; when x = 3, y = 3. The solutions are (0,0) and (3,3). A common error is forgetting that x = 0 is a valid solution when factoring out x - always include it when you factor out a variable.

Question 13

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

Explanation: We need to find how many times the parabola y = x² + 4x + 5 intersects the horizontal line y = 1. This system has a quadratic and a constant linear equation, solved by substitution. Setting equal: x² + 4x + 5 = 1, which simplifies to x² + 4x + 4 = 0. This factors as (x + 2)² = 0, giving x = -2 as a repeated root. Since we have one repeated solution, the line is tangent to the parabola at exactly one point. A common misconception is that a quadratic equation always has two distinct solutions, but when the discriminant equals zero, there's exactly one solution (the vertex touches the line).

Question 14

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

Explanation: We need to find where the parabola y = -x² + 4x + 4 intersects the horizontal line y = 4. This system has a quadratic and a constant linear equation, solved by substitution. Setting equal: -x² + 4x + 4 = 4, which simplifies to -x² + 4x = 0, then -x(x - 4) = 0. This gives x = 0 or x = 4. Since y = 4 for both solutions (given by the line equation), our intersection points are (0,4) and (4,4). A common error is thinking a horizontal line can only intersect a parabola once, but it can intersect 0, 1, or 2 times depending on the parabola's position and shape.

Question 15

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

Explanation: The question asks for the number of intersection points between y = x² - 2x - 3 and y = -x + 1, with the graphs already drawn showing visible intersections. This system combines a quadratic and linear equation, solved by substitution. Setting equal: x² - 2x - 3 = -x + 1, which becomes x² - x - 4 = 0. The discriminant is b² - 4ac = (-1)² - 4(1)(-4) = 1 + 16 = 17 > 0, confirming two real solutions. Since the discriminant is positive, the parabola and line intersect at exactly 2 points. When reading graphs on the SAT, trust the algebraic verification over visual estimation.

Question 16

1. $(1,3)$ only
2. $(0,1)$ and $(4,9)$ (correct answer)
3. $(0,1)$ and $(6,13)$
4. $(0,1)$ only

Explanation: The question asks for all solutions (x, y) to the system of equations modeling the ball's height and the platform's height, representing the times when the ball is exactly at the platform height. This system consists of a quadratic equation (downward-opening parabola) and a linear equation (line), which can intersect at 0, 1, or 2 points, and we solve by substitution, setting the y-expressions equal. Set -x² + 6x + 1 = 2x + 1, then subtract 2x + 1 from both sides to get -x² + 4x = 0; multiply by -1 for x² - 4x = 0, and factor as x(x - 4) = 0, yielding x = 0 or x = 4. Substitute into the line: for x = 0, y = 1; for x = 4, y = 9, so solutions are (0, 1) and (4, 9). A key error might be failing to multiply by -1, leading to incorrect roots, or forgetting to compute y-values. Another mistake is assuming only positive x-values are valid without context. For test-taking, since a line and parabola intersect at most twice, verify both solutions satisfy the original equations to confirm.

Question 17

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

Explanation: This problem asks how many times the line $y = x + 2$ intersects the parabola $y = x^2 + 2$. This is a linear-nonlinear system solved by substitution. Setting equal: $x + 2 = x^2 + 2$, which simplifies to $x = x^2$, or $x^2 - x = 0$, giving $x(x - 1) = 0$. Thus $x = 0$ or $x = 1$, yielding two distinct solutions. The intersection points are $(0, 2)$ and $(1, 3)$. Remember that a line and parabola can intersect at 0, 1, or 2 points, determined by the discriminant of the resulting quadratic.

Question 18

1. $(0,5)$ and $(2,1)$ (correct answer)
2. $(0,5)$ and $(2,5)$
3. $(1,3)$ and $(3,-1)$
4. $(1,3)$ and $(3,3)$

Explanation: We need all solutions where the line $y = -2x + 5$ meets the parabola $y = x^2 - 4x + 5$. This is a linear-nonlinear system solved by substitution. Setting equal: $-2x + 5 = x^2 - 4x + 5$, which simplifies to $0 = x^2 - 2x$, or $x(x - 2) = 0$, giving $x = 0$ or $x = 2$. When $x = 0$, $y = -2(0) + 5 = 5$; when $x = 2$, $y = -2(2) + 5 = 1$. The solutions are $(0, 5)$ and $(2, 1)$. Always verify by substituting back into both original equations to avoid calculation errors.

Question 19

1. $(1,-2)$ and $(7,21)$
2. $(0,-5)$ and $(7,21)$
3. $(1,-2)$ and $(7,16)$
4. $(0,-5)$ and $(7,16)$ (correct answer)

Explanation: We need to find where the profit curve ($y = x^2 - 4x - 5$) meets the break-even line ($y = 3x - 5$). This is a system with a parabola and a line, solved by substitution. Setting the equations equal: $x^2 - 4x - 5 = 3x - 5$, which simplifies to $x^2 - 7x = 0$, or $x(x - 7) = 0$, giving $x = 0$ or $x = 7$. When $x = 0$, $y = 3(0) - 5 = -5$; when $x = 7$, $y = 3(7) - 5 = 16$. The solutions are ($0, -5$) and ($7, 16$). A key error would be arithmetic mistakes when substituting back to find y-values.

Question 20

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

Explanation: The question asks for the sum x₁ + x₂ of intersection points between a drone's quadratic altitude and a linear safety ceiling. This quadratic-linear system is solved by substitution, recalling possible intersections of 0, 1, or 2 for a parabola and line. Set x² - 2x + 2 = 2x - 2, simplify to x² - 4x + 4 = 0, or (x - 2)² = 0, with double root x = 2. The sum of roots from the quadratic coefficients is 4, treating the tangent case as repeated roots. Errors often occur by counting only one root and ignoring multiplicity for sums. A strategy is to use Vieta's formulas directly for sums without solving fully.