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Difficulty: MediumSystems of Linear and Non-Linear Equations

In the standard (x,y)(x, y) coordinate plane, the graph of the parabola y=x2−5x+2y = x^2 - 5x + 2 intersects the line 2x−y=42x - y = 4 at two points, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). What is the value of x1x2+y1y2x_1 x_2 + y_1 y_2?

  1. A
    50
  2. B
    22
  3. -10Answer
  4. D
    6
  5. E
    102

Answer

The value of x1x2+y1y2x_1 x_2 + y_1 y_2 is −10-10.
The correct answer is −10-10. To find the intersection points, substitute the linear equation y=2x−4y = 2x - 4 into the quadratic equation y=x2−5x+2y = x^2 - 5x + 2, which simplifies to x2−7x+6=0x^2 - 7x + 6 = 0. Factoring gives (x−1)(x−6)=0(x-1)(x-6) = 0, yielding xx-coordinates x1=1x_1 = 1 and x2=6x_2 = 6. Substituting these back into the linear equation gives the corresponding yy-coordinates y1=−2y_1 = -2 and y2=8y_2 = 8. Evaluating the requested expression gives (1)(6)+(−2)(8)=6−16=−10(1)(6) + (-2)(8) = 6 - 16 = -10.

Step-by-Step Solution

1
Isolate yy in the linear equation.
y=2x−4y = 2x - 4
This allows for substitution into the quadratic equation.
2
Substitute y=2x−4y = 2x - 4 into the quadratic equation y=x2−5x+2y = x^2 - 5x + 2.
2x−4=x2−5x+22x - 4 = x^2 - 5x + 2
Setting the two equations equal to each other helps find the x-coordinates of the intersection points.
3
Rearrange the equation into standard quadratic form.
x2−7x+6=0x^2 - 7x + 6 = 0
This sets up the equation for factoring.
4
Factor the quadratic equation.
(x−1)(x−6)=0(x-1)(x-6) = 0
Factoring reveals the solutions for the x-coordinates.
5
Solve for the two x-coordinates.
x1=1x_1 = 1 and x2=6x_2 = 6
These are the x-coordinates of the two intersection points.
6
Find the corresponding y-coordinates by substituting the x-values back into the linear equation.
For x1=1x_1 = 1: y1=2(1)−4=−2y_1 = 2(1) - 4 = -2. For x2=6x_2 = 6: y2=2(6)−4=8y_2 = 2(6) - 4 = 8.
This gives the full coordinate pairs: (1,−2)(1, -2) and (6,8)(6, 8).
7
Calculate the value of the expression x1x2+y1y2x_1 x_2 + y_1 y_2.
(1)(6)+(−2)(8)=6−16=−10(1)(6) + (-2)(8) = 6 - 16 = -10
This provides the final required value.

Key Concept

Solving systems of linear and quadratic equations by substitution and evaluating expressions of their coordinate solutions.
Estimated Time:1m 30s
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Topics

Systems of Linear and Non-Linear Equations

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