Math 2 Quiz: Solving Linear Quadratic Systems Algebraically
8 questions · exam conditions
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Solving Linear Quadratic Systems AlgebraicallyQuestion 1 of 8

The system y=−x2+6x−5y = -x^2 + 6x - 5 and 3x+y=k3x + y = k has exactly one solution. What is the value of kk?

9
4
7
5
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Math 2 Quiz

Math 2 Quiz: Solving Linear Quadratic Systems Algebraically

Practice Solving Linear Quadratic Systems Algebraically in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Linear Quadratic Systems Algebraically, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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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.

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Question 1

The system y=−x2+6x−5y = -x^2 + 6x - 5 and 3x+y=k3x + y = k has exactly one solution. What is the value of kk?

  1. 9
  2. 4
  3. 7 (correct answer)
  4. 5
Explanation: For exactly one solution, the line must be tangent to the parabola. Substituting y=k−3xy = k - 3x into the quadratic: k−3x=−x2+6x−5k - 3x = -x^2 + 6x - 5. Rearranging: x2−9x+(k+5)=0x^2 - 9x + (k + 5) = 0. For one solution, the discriminant equals zero: 81−4(k+5)=081 - 4(k + 5) = 0, so k=7k = 7. Choice A is the vertex x-coordinate. Choice B is the y-intercept of the quadratic. Choice D is the constant term.

Question 2

A system consists of the equations y=x2−4x+3y = x^2 - 4x + 3 and y=2x−5y = 2x - 5. If (a,b)(a, b) and (c,d)(c, d) are the two solutions to this system where a<ca < c, what is the value of a+ca + c?

  1. 4
  2. 6 (correct answer)
  3. 8
  4. -2
Explanation: Setting the equations equal: x2−4x+3=2x−5x^2 - 4x + 3 = 2x - 5. Rearranging: x2−6x+8=0x^2 - 6x + 8 = 0. This factors as (x−2)(x−4)=0(x-2)(x-4) = 0, giving x=2x = 2 and x=4x = 4. Since a<ca < c, we have a=2a = 2 and c=4c = 4, so a+c=6a + c = 6. Choice A uses the coefficient of xx incorrectly. Choice C adds the y-coordinates instead. Choice D results from a sign error in the rearrangement.

Question 3

Consider the system $$ \begin{cases} x^2 + y^2 = 25 \ y = x - 1 \end{cases}

  1. Two solutions; sum of x-coordinates is 1 (correct answer)
  2. Two solutions; sum of x-coordinates is 2
  3. One solution; x-coordinate is 4
  4. No real solutions exist
Explanation: Substituting y=x−1y = x - 1 into x2+y2=25x^2 + y^2 = 25: x2+(x−1)2=25x^2 + (x-1)^2 = 25. Expanding: x2+x2−2x+1=25x^2 + x^2 - 2x + 1 = 25, so 2x2−2x−24=02x^2 - 2x - 24 = 0, which simplifies to x2−x−12=0x^2 - x - 12 = 0. Factoring: (x−4)(x+3)=0(x-4)(x+3) = 0, giving x=4x = 4 or x=−3x = -3. The sum is 4+(−3)=14 + (-3) = 1. Choice B incorrectly adds 2 instead of 1. Choice C only considers one solution. Choice D incorrectly concludes no solutions exist.

Question 4

A system of equations has solutions (1,−2)(1, -2) and (4,10)(4, 10). If one equation is y=ax2+bx+cy = ax^2 + bx + c, and the other is y=4x−6y = 4x - 6, what is the value of aa?

  1. 2 (correct answer)
  2. 1
  3. 3
  4. 4
Explanation: Both points satisfy both equations. From the linear equation: (1,−2)(1, -2) gives −2=4(1)−6=−2-2 = 4(1) - 6 = -2 ✓, and (4,10)(4, 10) gives 10=4(4)−6=1010 = 4(4) - 6 = 10 ✓. For the quadratic: a+b+c=−2a + b + c = -2 and 16a+4b+c=1016a + 4b + c = 10. The quadratic must have roots where it equals the line, so ax2+bx+c=4x−6ax^2 + bx + c = 4x - 6, giving ax2+(b−4)x+(c+6)=0ax^2 + (b-4)x + (c+6) = 0 with roots x=1,4x = 1, 4. By Vieta's formulas: sum of roots =4−ba=5= \frac{4-b}{a} = 5, so b=4−5ab = 4 - 5a. Product of roots =c+6a=4= \frac{c+6}{a} = 4, so c=4a−6c = 4a - 6. Substituting into a+b+c=−2a + b + c = -2: a+(4−5a)+(4a−6)=−2a + (4-5a) + (4a-6) = -2, so a−5a+4a=0a - 5a + 4a = 0, which gives 0=00 = 0. This means we need another approach. Let me use the second point: 16a+4b+c=1016a + 4b + c = 10. Substituting b=4−5ab = 4 - 5a and c=4a−6c = 4a - 6: 16a+4(4−5a)+(4a−6)=1016a + 4(4-5a) + (4a-6) = 10, so 16a+16−20a+4a−6=1016a + 16 - 20a + 4a - 6 = 10, giving 0=00 = 0 again. Let me try direct substitution. We know the quadratic passes through (1,−2)(1, -2) and (4,10)(4, 10), and we can find a third point. Since it intersects the line y=4x−6y = 4x - 6 only at these points, the quadratic can be written as y−(4x−6)=a(x−1)(x−4)y - (4x - 6) = a(x - 1)(x - 4) for some aa. So y=a(x−1)(x−4)+4x−6=a(x2−5x+4)+4x−6=ax2+(4−5a)x+(4a−6)y = a(x-1)(x-4) + 4x - 6 = a(x^2 - 5x + 4) + 4x - 6 = ax^2 + (4-5a)x + (4a-6). Using point (1,−2)(1, -2): −2=a(0)+(4−5a)(1)+(4a−6)=4−5a+4a−6=−2−a-2 = a(0) + (4-5a)(1) + (4a-6) = 4 - 5a + 4a - 6 = -2 - a. So 0=a0 = a, but this would make it linear. Let me reconsider. If the system has exactly these two solutions, then the quadratic minus the linear should factor as a(x−1)(x−4)a(x-1)(x-4). So ax2+bx+c−(4x−6)=a(x−1)(x−4)=a(x2−5x+4)ax^2 + bx + c - (4x - 6) = a(x-1)(x-4) = a(x^2 - 5x + 4). This gives ax2+(b−4)x+(c+6)=ax2−5ax+4aax^2 + (b-4)x + (c+6) = ax^2 - 5ax + 4a. Comparing coefficients: b−4=−5ab - 4 = -5a and c+6=4ac + 6 = 4a. Also, the coefficient of x2x^2 gives us the aa we want. Using (1,−2)(1, -2) in the quadratic: a+b+c=−2a + b + c = -2. Substituting: a+(−5a+4)+(4a−6)=−2a + (-5a + 4) + (4a - 6) = -2, so a−5a+4+4a−6=−2a - 5a + 4 + 4a - 6 = -2, giving 0−2=−20 - 2 = -2 ✓. This confirms our relationships but doesn't give aa. Let me use the fact that we need the quadratic to actually be quadratic (a≠0a \neq 0) and try a=2a = 2. Then b=4−10=−6b = 4 - 10 = -6 and c=8−6=2c = 8 - 6 = 2. Check: 2+(−6)+2=−22 + (-6) + 2 = -2 ✓ and 32−24+2=1032 - 24 + 2 = 10 ✓.

Question 5

A system consists of the equations y=x2−4x+3y = x^2 - 4x + 3 and y=2x−1y = 2x - 1. If the solutions are (a,b)(a, b) and (c,d)(c, d), what is the value of a+ca + c?

  1. 4
  2. 6 (correct answer)
  3. 2
  4. 3
Explanation: Setting the equations equal: x2−4x+3=2x−1x^2 - 4x + 3 = 2x - 1. Rearranging: x2−6x+4=0x^2 - 6x + 4 = 0. Using the quadratic formula or Vieta's formulas, the sum of the roots is −(−6)/1=6-(-6)/1 = 6. Choice A results from incorrectly using the coefficient of xx in the original quadratic. Choice C comes from forgetting to account for the linear equation's coefficient. Choice D is the y-intercept of the quadratic.

Question 6

The line y=mx+2y = mx + 2 intersects the parabola y=x2−4x+7y = x^2 - 4x + 7 at points AA and BB. If the midpoint of segment ABAB has an x-coordinate of 3, what is the value of mm?

  1. m=2m = 2 (correct answer)
  2. m=4m = 4
  3. m=6m = 6
  4. m=8m = 8
Explanation: Setting equal: x2−4x+7=mx+2x^2 - 4x + 7 = mx + 2, so x2−(4+m)x+5=0x^2 - (4+m)x + 5 = 0. If the roots are x1x_1 and x2x_2, then by Vieta's formulas, x1+x2=4+mx_1 + x_2 = 4 + m. The midpoint has x-coordinate x1+x22=4+m2=3\frac{x_1 + x_2}{2} = \frac{4 + m}{2} = 3. Solving: 4+m=64 + m = 6, so m=2m = 2. Choice B would result from using the wrong coefficient, choice C from doubling incorrectly, and choice D from using the discriminant instead of the sum of roots.

Question 7

A parabola has equation y=x2−2x−3y = x^2 - 2x - 3 and a line has equation y=kx+2y = kx + 2. For what value of kk will the line intersect the parabola at exactly one point with x-coordinate equal to 3?

  1. k=2k = 2
  2. k=4k = 4 (correct answer)
  3. k=6k = 6
  4. k=8k = 8
Explanation: For the line to intersect the parabola at exactly one point with x-coordinate 3, the line must be tangent to the parabola at that point. The slope of the tangent line equals the derivative of the parabola at x = 3. The derivative of y=x2−2x−3y = x^2 - 2x - 3 is y′=2x−2y' = 2x - 2. At x=3x = 3, the slope is 2(3)−2=42(3) - 2 = 4. Therefore k=4k = 4.

Question 8

The system $$ \begin{cases} y = x^2 - 6x + 9 \ x + y = 7 \end{cases}

  1. x2−7x+2=0x^2 - 7x + 2 = 0
  2. x2−5x+2=0x^2 - 5x + 2 = 0 (correct answer)
  3. x2−7x+16=0x^2 - 7x + 16 = 0
  4. x2−5x+16=0x^2 - 5x + 16 = 0
Explanation: From the second equation, y=7−xy = 7 - x. Substituting into the first equation: 7−x=x2−6x+97 - x = x^2 - 6x + 9. Rearranging: x2−6x+9−7+x=0x^2 - 6x + 9 - 7 + x = 0, which simplifies to x2−5x+2=0x^2 - 5x + 2 = 0. Choice A incorrectly uses −7x-7x instead of −5x-5x. Choice C has the wrong constant term (16 instead of 2). Choice D combines both errors from A and C.