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Difficulty: Very hardQuadratic Equations and Factoring

Consider the quadratic equation x2−2(k+1)x+(k2+4k+1)=0x^2 - 2(k + 1)x + (k^2 + 4k + 1) = 0, where kk is a real constant. Let α\alpha and β\beta be the real roots of this equation. If α2+β2=18\alpha^2 + \beta^2 = 18, what is the sum of all possible values of kk?

  1. −22-2\sqrt{2}Answer
  2. B
    00
  3. C
    222\sqrt{2}
  4. D
    2−232 - 2\sqrt{3}
  5. E
    −42-4\sqrt{2}

Answer

The sum of all possible real values of kk is −22-2\sqrt{2}.
Using Vieta's formulas, α+β=2(k+1)\alpha + \beta = 2(k + 1) and αβ=k2+4k+1\alpha\beta = k^2 + 4k + 1. Expanding α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta gives 2k2+22k^2 + 2. Setting 2k2+2=182k^2 + 2 = 18 yields k2=8k^2 = 8, so k=±22k = \pm 2\sqrt{2}. Furthermore, for the roots α\alpha and β\beta to be real numbers, the discriminant Δ=[−2(k+1)]2−4(k2+4k+1)=−8k\Delta = [-2(k+1)]^2 - 4(k^2+4k+1) = -8k must be non-negative, requiring k≤0k \le 0. Only k=−22k = -2\sqrt{2} satisfies this requirement. Hence, the sum of all valid values of kk is −22-2\sqrt{2}.

Step-by-Step Solution

1
Apply Vieta's formulas to express the sum and product of roots in terms of kk.
α+β=2(k+1)\alpha + \beta = 2(k + 1) and αβ=k2+4k+1\alpha\beta = k^2 + 4k + 1.
For any quadratic equation x2−bx+c=0x^2 - bx + c = 0, the sum of roots is bb and the product of roots is cc.
2
Express α2+β2\alpha^2 + \beta^2 in terms of kk using the algebraic identity (α+β)2−2αβ(\alpha + \beta)^2 - 2\alpha\beta.
α2+β2=[2(k+1)]2−2(k2+4k+1)=4(k2+2k+1)−2k2−8k−2=2k2+2\alpha^2 + \beta^2 = [2(k + 1)]^2 - 2(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 2k^2 - 8k - 2 = 2k^2 + 2.
Expanding and simplifying allows us to relate the given sum of squares to the parameter kk.
3
Set the simplified expression equal to 1818 and solve for kk.
2k2+2=18  ⟹  2k2=16  ⟹  k2=8  ⟹  k=±222k^2 + 2 = 18 \implies 2k^2 = 16 \implies k^2 = 8 \implies k = \pm 2\sqrt{2}.
This yields two candidate real values for the parameter kk.
4
Analyze the discriminant Δ\Delta to enforce the condition that roots α\alpha and β\beta are real numbers.
Δ=[−2(k+1)]2−4(1)(k2+4k+1)=4(k2+2k+1)−4(k2+4k+1)=−8k\Delta = [-2(k + 1)]^2 - 4(1)(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 4(k^2 + 4k + 1) = -8k. For real roots, Δ≥0  ⟹  −8k≥0  ⟹  k≤0\Delta \ge 0 \implies -8k \ge 0 \implies k \le 0.
A quadratic equation has real roots if and only if its discriminant is non-negative.
5
Filter candidate values of kk against the condition k≤0k \le 0.
k=−22k = -2\sqrt{2} satisfies k≤0k \le 0 (yielding Δ=162>0\Delta = 16\sqrt{2} > 0), while k=22k = 2\sqrt{2} fails k≤0k \le 0 (yielding Δ=−162<0\Delta = -16\sqrt{2} < 0). Thus, k=−22k = -2\sqrt{2} is the unique valid value.
Values of kk that make the discriminant negative produce non-real complex roots and must be discarded as extraneous.

Key Concept

Quadratic Vieta's formulas combined with discriminant non-negativity constraint for real roots
Estimated Time:2m 30s
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Topics

Quadratic Equations and Factoring

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