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Difficulty: Very hardQuadratic Equations and Factoring

The quadratic equation x2−(k−2)x+(k−5)=0x^2 - (k - 2)x + (k - 5) = 0 has two distinct real roots, α\alpha and β\beta. If 1α2+1β2=1\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 1, what is the value of the constant kk?

  1. A
    94\frac{9}{4}
  2. 114\frac{11}{4}Answer
  3. C
    72\frac{7}{2}
  4. D
    318\frac{31}{8}
  5. E
    314\frac{31}{4}

Answer

The value of the constant kk is 114\frac{11}{4}.
Using Vieta's formulas for the quadratic equation x2−(k−2)x+(k−5)=0x^2 - (k - 2)x + (k - 5) = 0, we have α+β=k−2\alpha + \beta = k - 2 and αβ=k−5\alpha\beta = k - 5. The given condition 1α2+1β2=1\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 1 simplifies to (α+β)2−2αβ(αβ)2=1\frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 1. Substituting the expressions in terms of kk yields (k−2)2−2(k−5)(k−5)2=1\frac{(k - 2)^2 - 2(k - 5)}{(k - 5)^2} = 1, which expands to k2−6k+14=k2−10k+25k^2 - 6k + 14 = k^2 - 10k + 25. Subtracting k2k^2 from both sides gives 4k=114k = 11, so k=114k = \frac{11}{4}.

Step-by-Step Solution

1
Apply Vieta's formulas to express root sum and product in terms of kk
α+β=k−2\alpha + \beta = k - 2 and αβ=k−5\alpha\beta = k - 5
For any standard quadratic ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is −ba-\frac{b}{a} and the product of roots is ca\frac{c}{a}.
2
Rewrite the given sum of reciprocal squares using algebraic identities
\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}
Combining fractions over a common denominator allows substitution of the known sum α+β\alpha+\beta and product αβ\alpha\beta.
3
Substitute the expressions for sum and product into the equation
\frac{(k - 2)^2 - 2(k - 5)}{(k - 5)^2} = 1
Set the algebraic expression equal to the given target value of 1.
4
Expand both numerator and denominator and solve for kk
\frac{k^2 - 4k + 4 - 2k + 10}{k^2 - 10k + 25} = 1 \implies k^2 - 6k + 14 = k^2 - 10k + 25 \implies 4k = 11 \implies k = \frac{11}{4}
Equating numerator and denominator eliminates the quadratic k2k^2 terms, yielding a linear equation in kk.
5
Verify that k=114k = \frac{11}{4} yields real, distinct, non-zero roots
Discriminant D=(k−2)2−4(k−5)=0.5625+9=9.5625>0D = (k-2)^2 - 4(k-5) = 0.5625 + 9 = 9.5625 > 0, and αβ=−2.25≠0\alpha\beta = -2.25 \neq 0
Ensures the quadratic has two distinct real roots as required by the problem prompt.

Key Concept

Vieta's Formulas and Symmetric Polynomial Transformations of Quadratic Roots
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Topics

Quadratic Equations and Factoring

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