Algebra · Lesson 6

Linear-quadratic systems

Substitute, then solve a quadratic. When one equation is a line and the other is a parabola, both are usually solved for yy. Set the right sides equal, move every term to one side, and solve the quadratic for xx. Then put each xx into the line to get yy. For y=x2−3xy = x^2 - 3x and y=x+5y = x + 5: x2−4x−5=0x^2 - 4x - 5 = 0, so x=5x = 5 or x=−1x = -1, and the points are (5,10)(5, 10) and (−1,4)(-1, 4).

Count with the discriminant. A line meets a parabola in 0, 1 or 2 points. After you combine the equations into ax2+bx+c=0ax^2 + bx + c = 0, the discriminant b2−4acb^2 - 4ac tells you how many: positive means two points, zero means one point (the line is tangent), negative means none. To find a constant that makes the line tangent, set the discriminant equal to 0 and solve.

Circles work the same way. Substitute the line into the circle equation, expand any squared binomial in full, and solve. Then reread the question: it may ask for xx, for yy or for the whole point.

Worked example Easy

The graphs of y = x^2 - 3x and y = x + 5−3−2−11234567−4−224681012140xyPQ
The graphs of y=x2−3xy = x^2 - 3x and y=x+5y = x + 5 An upward-opening parabola that crosses the x-axis at x = 0 and x = 3 and has its lowest point at (1.5, −2.25), and a rising line with slope 1 and y-intercept 5. The graphs cross at two labelled points: P, to the left of the y-axis, and Q, in the first quadrant.

The graphs of y=x2−3xy = x^2 - 3x and y=x+5y = x + 5 are shown in the standard (x, y) coordinate plane. They intersect at points P and Q. What are the coordinates of P and Q?

  1. (−5,0)(-5, 0) and (1,6)(1, 6)
  2. (−1,−2)(-1, -2) and (5,10)(5, 10)
  3. (−1,4)(-1, 4) and (5,10)(5, 10)Answer
  4. (0,5)(0, 5) and (3,8)(3, 8)

How to solve it

  1. Set the right sides equal: x2−3x=x+5x^2 - 3x = x + 5.
  2. Move every term to the left: x2−4x−5=0x^2 - 4x - 5 = 0, so (x−5)(x+1)=0(x - 5)(x + 1) = 0.
  3. The roots are x=−1x = -1 and x=5x = 5.
  4. Use the line for yy: −1+5=4-1 + 5 = 4 and 5+5=105 + 5 = 10. The points are (−1,4)(-1, 4) and (5,10)(5, 10).

Why each choice is right or wrong

  • A. Incorrect. (−5,0)(-5, 0) and (1,6)(1, 6) come from factoring x2−4x−5x^2 - 4x - 5 as (x+5)(x−1)(x + 5)(x - 1). Its factors are (x−5)(x+1)(x - 5)(x + 1), so the roots are 5 and −1-1.
  • B. Incorrect. (−1,−2)(-1, -2) evaluates the parabola at x=−1x = -1 as 1−31 - 3, treating −3(−1)-3(-1) as −3-3. In fact (−1)2−3(−1)=4(-1)^2 - 3(-1) = 4, which matches the line.
  • C. Correct. x2−3x=x+5x^2 - 3x = x + 5 gives (x−5)(x+1)=0(x - 5)(x + 1) = 0. At x=−1x = -1, y=4y = 4; at x=5x = 5, y=10y = 10. Point P is (−1,4)(-1, 4) and Q is (5,10)(5, 10).
  • D. Incorrect. (0,5)(0, 5) and (3,8)(3, 8) use the zeros of x2−3xx^2 - 3x, where the parabola meets the x-axis, not where it meets the line. Set the two expressions equal to each other instead.

Worked example Medium

Which of the following gives all the solutions (x,y)(x, y) of the system x2+y2=25x^2 + y^2 = 25 and y=x+1y = x + 1?

  1. (−4,−3)(-4, -3) and (3,4)(3, 4)Answer
  2. (−23,1−23)(-2\sqrt{3}, 1 - 2\sqrt{3}) and (23,1+23)(2\sqrt{3}, 1 + 2\sqrt{3})
  3. (−3,−2)(-3, -2) and (4,5)(4, 5)
  4. (3,4)(3, 4) only

How to solve it

  1. Substitute y=x+1y = x + 1: x2+(x+1)2=25x^2 + (x + 1)^2 = 25.
  2. Expand in full: x2+x2+2x+1=25x^2 + x^2 + 2x + 1 = 25, so 2x2+2x−24=02x^2 + 2x - 24 = 0.
  3. Divide by 2 and factor: x2+x−12=(x+4)(x−3)=0x^2 + x - 12 = (x + 4)(x - 3) = 0, so x=−4x = -4 or x=3x = 3.
  4. Use the line: y=−3y = -3 and y=4y = 4. The solutions are (−4,−3)(-4, -3) and (3,4)(3, 4).

Why each choice is right or wrong

  • A. Correct. x2+(x+1)2=25x^2 + (x + 1)^2 = 25 simplifies to x2+x−12=0x^2 + x - 12 = 0, so x=−4x = -4 or x=3x = 3, and y=x+1y = x + 1 gives (−4,−3)(-4, -3) and (3,4)(3, 4). Both points satisfy 16+9=2516 + 9 = 25 and 9+16=259 + 16 = 25.
  • B. Incorrect. The 232\sqrt{3} values come from expanding (x+1)2(x + 1)^2 as x2+1x^2 + 1, which drops the middle term 2x2x and gives 2x2=242x^2 = 24.
  • C. Incorrect. (−3,−2)(-3, -2) and (4,5)(4, 5) come from factoring x2+x−12x^2 + x - 12 as (x+3)(x−4)(x + 3)(x - 4). That product is x2−x−12x^2 - x - 12, so the signs are swapped.
  • D. Incorrect. (3,4)(3, 4) only keeps the positive root. The point (−4,−3)(-4, -3) also lies on both graphs: 16+9=2516 + 9 = 25 and −4+1=−3-4 + 1 = -3.

Trap Answering with x when the question asks for y

Solving the quadratic gives the xx-coordinates, and a distractor usually lists exactly those. Before you choose, reread the question. If it asks for yy or for the points, put each xx into the simpler equation, usually the line.

Trap Sign slips when you move terms

Moving x+5x + 5 to the left side changes both signs: x2−3x−x−5=0x^2 - 3x - x - 5 = 0. A slip here changes the factors and every answer after it. In the discriminant, keep the minus in b2−4acb^2 - 4ac and use the whole constant term cc.

Calculator When a calculator is faster

A graph is fast for these systems. With the built-in Desmos calculator if you test online, type both equations and click each crossing point to read its coordinates. On a graphing calculator, use the intersect feature once for each point.

For a tangent-line question with a constant kk, add a slider for kk and watch the line touch the parabola, then confirm the exact value with the discriminant. On a four-function calculator, solve by substitution and test the choices.