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Zorluk: Çok zorQuadratic Equations and Polynomial Factoring

For how many integer values of kk does the quadratic equation (k−2)x2−2(k+1)x+(k+3)=0(k - 2)x^2 - 2(k + 1)x + (k + 3) = 0 have two distinct real roots α\alpha and \beta suchthat such that \frac{1}{\alpha} + \frac{1}{\beta} < 1$?

  1. A
    2
  2. 3Cevap
  3. C
    5
  4. D
    6
  5. E
    7

Cevap

There are exactly 3 integer values of kk that satisfy all given conditions.
To find the number of valid integer values for kk, we apply the condition for distinct real roots (D>0D > 0), which yields k>−7k > -7 with k≠2k \neq 2. Expressing the sum of reciprocals as α+βαβ=2(k+1)k+3<1\frac{\alpha + \beta}{\alpha\beta} = \frac{2(k+1)}{k+3} < 1, solving k−1k+3<0\frac{k-1}{k+3} < 0 yields −3<k<1-3 < k < 1. The integer values within this range are −2,−1,-2, -1, and 00, giving a total of 3 valid integer values.

Adım Adım Çözüm

1
Ensure the given equation is a valid quadratic equation and determine the condition for two distinct real roots.
For the equation to be quadratic, the coefficient of x2x^2 must be non-zero, so k−2≠0  ⟹  k≠2k - 2 \neq 0 \implies k \neq 2. For distinct real roots, the discriminant D>0D > 0. D=[−2(k+1)]2−4(k−2)(k+3)=4(k2+2k+1)−4(k2+k−6)=4(k+7)D = [-2(k+1)]^2 - 4(k-2)(k+3) = 4(k^2 + 2k + 1) - 4(k^2 + k - 6) = 4(k + 7). Thus, 4(k+7)>0  ⟹  k>−74(k+7) > 0 \implies k > -7.
A quadratic equation has two distinct real roots if and only if its discriminant is strictly positive and the leading coefficient is non-zero.
2
Apply Vieta's formulas to express the sum and product of the roots in terms of kk.
\alpha + \beta = \frac{2(k+1)}{k-2}and and \alpha\beta = \frac{k+3}{k-2}.Notethatfor. Note that for \frac{1}{\alpha} + \frac{1}{\beta}toexist,neitherrootcanbe0,whichrequires to exist, neither root can be 0, which requires \alpha\beta
eq 0 \implies k
eq -3$.
Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots.
3
Simplify the condition 1α+1β<1\frac{1}{\alpha} + \frac{1}{\beta} < 1 using Vieta's relations.
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{\frac{2(k+1)}{k-2}}{\frac{k+3}{k-2}} = \frac{2(k+1)}{k+3}.Weneed. We need \frac{2k+2}{k+3} < 1$.
Combining the fractions over a common denominator allows direct substitution of the root sum and product.
4
Solve the algebraic inequality 2k+2k+3<1\frac{2k+2}{k+3} < 1.
\frac{2k+2}{k+3} - 1 < 0 \implies \frac{(2k+2) - (k+3)}{k+3} < 0 \implies \frac{k-1}{k+3} < 0 .Thefractionisnegativewhenthenumeratoranddenominatorhaveoppositesigns,whichgivestheinterval. The fraction is negative when the numerator and denominator have opposite signs, which gives the interval -3 < k < 1$.
Subtracting 1 to get 0 on one side prevents sign errors that occur when cross-multiplying variable expressions.
5
Combine all constraints on kk and count the valid integer values.
Constraints on kk: (1) k>−7k > -7, (2) k≠2k \neq 2, (3) k≠−3k \neq -3, and (4) −3<k<1-3 < k < 1. Intersecting these gives −3<k<1-3 < k < 1. The integer values in this open interval are k=−2,−1,0k = -2, -1, 0. Total count = 3.
All structural and domain restrictions must be satisfied simultaneously.

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