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Question

Difficulty: EasyQuadratic Equations

In the quadratic equation x2−kx+9=0x^2 - kx + 9 = 0, kk is a positive constant. If the equation has exactly one real solution, what is the value of kk?

Answer: 6

Answer

6
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 to have exactly one real solution, its discriminant must equal zero (b2−4ac=0b^2 - 4ac = 0). In the equation x2−kx+9=0x^2 - kx + 9 = 0, the coefficients are a=1a = 1, b=−kb = -k, and c=9c = 9. Setting the discriminant to zero gives (−k)2−4(1)(9)=0(-k)^2 - 4(1)(9) = 0, which simplifies to k2−36=0k^2 - 36 = 0. Solving this equation yields k2=36k^2 = 36, so k=6k = 6 or k=−6k = -6. Since kk is a positive constant, we reject the negative root, leaving k=6k = 6.

Step-by-Step Solution

1
Identify the condition for a quadratic equation to have exactly one real solution.
The discriminant of the quadratic equation must be equal to 0, which is represented by the formula b2−4ac=0b^2 - 4ac = 0.
The discriminant determines the number of real solutions of a quadratic equation. If the discriminant is 0, there is exactly one real solution.
2
Identify the coefficients aa, bb, and cc of the given equation x2−kx+9=0x^2 - kx + 9 = 0.
a=1a = 1, b=−kb = -k, and c=9c = 9.
These coefficients are required to compute the value of the discriminant.
3
Set the discriminant equal to 0 and simplify the equation.
(−k)2−4(1)(9)=0⇒k2−36=0(-k)^2 - 4(1)(9) = 0 \Rightarrow k^2 - 36 = 0.
Substituting the coefficients into the discriminant formula sets up the algebraic relationship to find kk.
4
Solve for the positive constant kk.
k2=36⇒k=6k^2 = 36 \Rightarrow k = 6 or k=−6k = -6. Since kk must be positive, k=6k = 6.
Solving the equation gives two possible values, but the negative solution is discarded because the problem specifies kk is a positive constant.

Key Concept

Discriminant of a quadratic equation

Alternative Method

Alternatively, a quadratic equation has exactly one real solution if it can be written as a perfect square trinomial in the form (x−d)2=0(x - d)^2 = 0, which expands to x2−2dx+d2=0x^2 - 2dx + d^2 = 0. Comparing this with x2−kx+9=0x^2 - kx + 9 = 0, we get d2=9d^2 = 9 and 2d=k2d = k. Since d2=9d^2 = 9, dd can be 33 or −3-3. Given that kk is positive and k=2dk = 2d, dd must also be positive, meaning d=3d = 3. Substituting this back gives k=2(3)=6k = 2(3) = 6.
Estimated Time:45s
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Topics

Quadratic Equations

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