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Zorluk: ZorQuadratic Equations

The quadratic equation x2−(k+3)x+(2k+3)=0x^2 - (k + 3)x + (2k + 3) = 0 has exactly one real solution, where kk is a constant. If k>1k > 1, what is the value of kk?

  1. A
    -1
  2. B
    1
  3. 3Cevap
  4. D
    5

Cevap

The value of kk is 33.
For the quadratic equation to have exactly one real solution, its discriminant must be equal to 00. Setting the discriminant of x2−(k+3)x+(2k+3)=0x^2 - (k + 3)x + (2k + 3) = 0 to 00 yields [−(k+3)]2−4(1)(2k+3)=0[-(k + 3)]^2 - 4(1)(2k + 3) = 0. Simplifying this yields k2−2k−3=0k^2 - 2k - 3 = 0, which factors as (k−3)(k+1)=0(k - 3)(k + 1) = 0. This gives two potential values for kk: 33 and −1-1. Because the problem specifies that k>1k > 1, the value of kk must be 33.

Adım Adım Çözüm

1
Identify the coefficients of the quadratic equation and state the condition for having exactly one real solution.
The coefficients of the quadratic equation x2−(k+3)x+(2k+3)=0x^2 - (k + 3)x + (2k + 3) = 0 are a=1a = 1, b=−(k+3)b = -(k + 3), and c=2k+3c = 2k + 3. For a quadratic equation to have exactly one real solution, its discriminant b2−4acb^2 - 4ac must equal 00.
A discriminant of zero indicates that the quadratic equation has two equal real roots, which is equivalent to exactly one distinct real solution.
2
Set up the discriminant equation in terms of kk and simplify it.
[−(k+3)]2−4(1)(2k+3)=0[-(k + 3)]^2 - 4(1)(2k + 3) = 0
(k2+6k+9)−(8k+12)=0(k^2 + 6k + 9) - (8k + 12) = 0
k2−2k−3=0k^2 - 2k - 3 = 0
This algebraic expansion and simplification transforms the discriminant condition into a solvable quadratic equation for kk.
3
Solve the quadratic equation k2−2k−3=0k^2 - 2k - 3 = 0 by factoring.
(k−3)(k+1)=0(k - 3)(k + 1) = 0, which gives k=3k = 3 or k=−1k = -1.
Factoring is the most direct method to find the roots of this quadratic equation.
4
Apply the constraint k>1k > 1 to select the correct value of kk.
Since kk must be greater than 11, the value k=−1k = -1 is discarded, leaving k=3k = 3.
The solution must satisfy all given constraints in the problem statement.

Anahtar Kavram

Discriminant of a Quadratic Equation

Alternatif Yöntem

Instead of solving the quadratic equation algebraically, you can test the given choices for kk. For the correct option of 33, substituting k=3k = 3 into the original equation yields x2−6x+9=0x^2 - 6x + 9 = 0. Factoring this expression gives (x−3)2=0(x - 3)^2 = 0, which clearly has exactly one real solution (x=3x = 3). Testing the other options would not produce a perfect square trinomial.
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Quadratic Equations

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