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Difficulty: HardSystems of Linear and Non-Linear Equations

In the standard (x,y)(x, y) coordinate plane, a parabola is defined by the equation y=x2+2x+7y = x^2 + 2x + 7 and a line is defined by the equation y=mx+3y = mx + 3, where mm is a constant. If the system of equations consisting of this parabola and line has exactly one real solution, and this solution lies in the first quadrant, what is the value of mm?

  1. A
    -2
  2. B
    -6
  3. 6Answer
  4. D
    232\sqrt{3}
  5. E
    2+272 + 2\sqrt{7}

Answer

6
The correct answer is 66. When setting the equations of the parabola and line equal, we get x2+(2−m)x+4=0x^2 + (2-m)x + 4 = 0. Setting the discriminant to zero yields (2−m)2−16=0(2-m)^2 - 16 = 0, which gives m=6m = 6 or m=−2m = -2. Substituting m=6m = 6 back gives a single intersection point of (2,15)(2, 15), which is in the first quadrant since both coordinates are positive. The other value, m=−2m = -2, gives an intersection point of (−2,7)(-2, 7), which is in the second quadrant.

Step-by-Step Solution

1
Set the equations of the parabola and the line equal to each other to find their intersection points.
x2+2x+7=mx+3x^2 + 2x + 7 = mx + 3
To find the coordinates where the two graphs intersect.
2
Rearrange the equation into standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
x2+(2−m)x+4=0x^2 + (2 - m)x + 4 = 0
To express the intersection condition as a single quadratic equation where we can analyze the number of solutions.
3
Set the discriminant of the quadratic equation to zero.
Δ=(2−m)2−4(1)(4)=0  ⟹  (2−m)2−16=0\Delta = (2 - m)^2 - 4(1)(4) = 0 \implies (2 - m)^2 - 16 = 0
For the system to have exactly one real solution, the quadratic equation must have a discriminant of zero (tangency).
4
Solve for the possible values of mm.
(2−m)2=16  ⟹  2−m=±4(2 - m)^2 = 16 \implies 2 - m = \pm 4, giving m=−2m = -2 or m=6m = 6.
To find all values of mm that result in exactly one intersection point.
5
Find the intersection point for each value of mm and determine which lies in the first quadrant.
For m=−2m = -2, (x+2)2=0  ⟹  x=−2(x+2)^2 = 0 \implies x = -2 and y=−2(−2)+3=7y = -2(-2)+3 = 7, yielding (−2,7)(-2, 7) (Quadrant II). For m=6m = 6, (x−2)2=0  ⟹  x=2(x-2)^2 = 0 \implies x = 2 and y=6(2)+3=15y = 6(2)+3 = 15, yielding (2,15)(2, 15) (Quadrant I).
To satisfy the condition that the single solution must lie in the first quadrant (where both x>0x > 0 and y>0y > 0).

Key Concept

Solving systems of linear and quadratic equations and applying the discriminant to find conditions for tangency.
Estimated Time:2m 0s
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Topics

Systems of Linear and Non-Linear Equations

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