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Difficulty: HardNonlinear Systems of Equations
y=2x2−kx+8y=4x−10\begin{aligned} y &= 2x^2 - kx + 8 \\ y &= 4x - 10 \end{aligned}

In the system of equations above, kk is a constant. If the system has exactly one real solution, what is the product of all possible values of kk?

  1. A
    -48
  2. -128Answer
  3. C
    -20
  4. D
    128

Answer

-128
To find the number of solutions to the system, set the two equations equal to each other: 2x2−kx+8=4x−102x^2 - kx + 8 = 4x - 10. Moving all terms to one side gives 2x2−(k+4)x+18=02x^2 - (k + 4)x + 18 = 0. For this quadratic equation to have exactly one real solution, its discriminant, Δ=B2−4AC\Delta = B^2 - 4AC, must equal zero. Substituting the coefficients A=2A = 2, B=−(k+4)B = -(k + 4), and C=18C = 18 into the discriminant formula gives (−(k+4))2−4(2)(18)=0(-(k + 4))^2 - 4(2)(18) = 0, which simplifies to (k+4)2−144=0(k + 4)^2 - 144 = 0. Solving this equation gives k+4=12  ⟹  k=8k + 4 = 12 \implies k = 8, or k+4=−12  ⟹  k=−16k + 4 = -12 \implies k = -16. The product of these possible values is 8×(−16)=−1288 \times (-16) = -128.

Step-by-Step Solution

1
Equate the quadratic and linear equations to find the condition for intersection.
2x2−kx+8=4x−102x^2 - kx + 8 = 4x - 10
To find the points where the two graphs intersect, their yy-values must be set equal to each other.
2
Rearrange the terms to form a standard quadratic equation of the form Ax2+Bx+C=0Ax^2 + Bx + C = 0.
2x2−(k+4)x+18=02x^2 - (k + 4)x + 18 = 0
Grouping terms allows us to determine the coefficients A=2A = 2, B=−(k+4)B = -(k + 4), and C=18C = 18 to use in the discriminant.
3
Set the discriminant Δ=B2−4AC\Delta = B^2 - 4AC of the quadratic equation to 00.
[−(k+4)]2−4(2)(18)=0  ⟹  (k+4)2−144=0[-(k + 4)]^2 - 4(2)(18) = 0 \implies (k + 4)^2 - 144 = 0
A system of a line and a parabola has exactly one real solution (meaning the line is tangent to the parabola) when the resulting quadratic equation has exactly one real root, which corresponds to a discriminant of zero.
4
Solve for the possible values of kk and calculate their product.
k+4=±12  ⟹  k=8k + 4 = \pm 12 \implies k = 8 or k=−16k = -16. The product is 8×(−16)=−1288 \times (-16) = -128.
Extracting the square root yields two possible values for kk, and multiplying them gives the requested product.

Key Concept

Using the discriminant of a quadratic equation to find parameter values that make a linear-quadratic system have exactly one real solution.

Alternative Method

Once the quadratic equation (k+4)2−144=0(k + 4)^2 - 144 = 0 is simplified to k2+8k−128=0k^2 + 8k - 128 = 0, Vieta's formulas can be applied directly to find the product of the possible values of kk. According to Vieta's formulas, the product of the roots of a quadratic equation ak2+bk+c=0ak^2 + bk + c = 0 is given by c/ac/a. Here, a=1a = 1 and c=−128c = -128, so the product of the roots is −128/1=−128-128 / 1 = -128.
Estimated Time:2m 0s
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Topics

Nonlinear Systems of Equations

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