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Zorluk: Çok zorQuadratic Equations and Factoring

For what values of the real constant kk does the quadratic equation (k−2)x2−2kx+(2k−3)=0(k-2)x^2 - 2kx + (2k - 3) = 0 have two distinct real roots r1r_1 and r2r_2 such that r1<1<r2r_1 < 1 < r_2?

  1. A
    1<k<61 < k < 6
  2. 2<k<52 < k < 5Cevap
  3. C
    k<2k < 2 or k>5k > 5
  4. D
    1<k<51 < k < 5
  5. E
    k>2k > 2

Cevap

The correct inequality expressing all possible values of kk is 2<k<52 < k < 5.
The correct inequality 2<k<52 < k < 5 is obtained by requiring (k−2)f(1)<0(k-2)f(1) < 0, which ensures x=1x = 1 falls between the two distinct real roots. Evaluating f(1)=k−5f(1) = k - 5 yields (k−2)(k−5)<0(k-2)(k-5) < 0, giving 2<k<52 < k < 5. The discriminant condition Δ=−4(k−1)(k−6)>0\Delta = -4(k-1)(k-6) > 0 gives 1<k<61 < k < 6, which fully encompasses (2,5)(2, 5).

Adım Adım Çözüm

1
Define the quadratic function and state the conditions for r1<1<r2r_1 < 1 < r_2.
Let f(x)=(k−2)x2−2kx+(2k−3)f(x) = (k-2)x^2 - 2kx + (2k - 3). For a quadratic function to have two real roots with x=1x = 1 located between them, the product of the leading coefficient (k−2)(k-2) and f(1)f(1) must be strictly negative, i.e., (k−2)f(1)<0(k-2)f(1) < 0.
If a parabola opens upwards (k−2>0k-2 > 0), its value at a point between its roots must be negative (f(1)<0f(1) < 0). If it opens downwards (k−2<0k-2 < 0), its value at a point between its roots must be positive (f(1)>0f(1) > 0).
2
Evaluate f(1)f(1) in terms of kk.
f(1)=(k−2)(1)2−2k(1)+(2k−3)=k−2−2k+2k−3=k−5f(1) = (k-2)(1)^2 - 2k(1) + (2k - 3) = k - 2 - 2k + 2k - 3 = k - 5.
Substitute x=1x = 1 directly into the expression for f(x)f(x).
3
Solve the inequality (k−2)f(1)<0(k-2)f(1) < 0.
(k−2)(k−5)<0  ⟹  2<k<5(k-2)(k-5) < 0 \implies 2 < k < 5.
The product of two linear factors (k−2)(k-2) and (k−5)(k-5) is negative between their roots, k=2k = 2 and k=5k = 5.
4
Verify discriminant condition Δ>0\Delta > 0 for real roots.
Δ=(−2k)2−4(k−2)(2k−3)=4k2−4(2k2−7k+6)=−4k2+28k−24=−4(k−1)(k−6)>0  ⟹  1<k<6\Delta = (-2k)^2 - 4(k-2)(2k-3) = 4k^2 - 4(2k^2 - 7k + 6) = -4k^2 + 28k - 24 = -4(k-1)(k-6) > 0 \implies 1 < k < 6.
Since the interval (2,5)(2, 5) is entirely contained within (1,6)(1, 6), any k∈(2,5)k \in (2, 5) automatically guarantees two distinct real roots.

Anahtar Kavram

Location of roots of quadratic equations and sign analysis of quadratic functions.
Tahmini Süre:2m 0s
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Konular

Quadratic Equations and Factoring

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