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Difficulty: HardQuadratic Equations

For which of the following values of kk does the equation x2−kx+9x−3=0\frac{x^2 - kx + 9}{x - 3} = 0 have exactly one real solution?

  1. A
    66
  2. B
    33
  3. −6-6Answer
  4. D
    −3-3

Answer

The correct value is −6-6.
For the equation x2−kx+9x−3=0\frac{x^2 - kx + 9}{x - 3} = 0 to have a solution, we must have x≠3x \neq 3. Multiplying both sides by the denominator yields the quadratic equation x2−kx+9=0x^2 - kx + 9 = 0. For a quadratic equation to have exactly one real solution, its discriminant must equal zero: (−k)2−4(1)(9)=k2−36=0(-k)^2 - 4(1)(9) = k^2 - 36 = 0, which gives k=6k = 6 or k=−6k = -6. If k=6k = 6, the equation is (x−3)2=0(x-3)^2 = 0, which has the single solution x=3x = 3. Since x=3x = 3 is excluded by the domain restriction, the equation has zero real solutions. If k=−6k = -6, the equation is (x+3)2=0(x+3)^2 = 0, which has the single solution x=−3x = -3. Since −3-3 is a valid value, this value of kk yields exactly one real solution.

Step-by-Step Solution

1
Identify the domain restriction of the rational equation.
The denominator x−3x - 3 cannot be zero, which means x≠3x \neq 3.
Division by zero is undefined, so x=3x = 3 cannot be a solution to the equation.
2
Simplify the equation by multiplying both sides by the denominator.
x2−kx+9=0x^2 - kx + 9 = 0 for x≠3x \neq 3.
A rational expression equals zero when its numerator is zero and its denominator is non-zero.
3
Set the discriminant of the quadratic equation to zero to find the values of kk that yield exactly one real root.
Δ=(−k)2−4(1)(9)=k2−36=0  ⟹  k=6\Delta = (-k)^2 - 4(1)(9) = k^2 - 36 = 0 \implies k = 6 or k=−6k = -6.
A quadratic equation has exactly one real solution (a double root) when its discriminant is zero.
4
Verify if the roots for each value of kk are valid under the domain restriction x≠3x \neq 3.
If k=6k = 6, x2−6x+9=(x−3)2=0  ⟹  x=3x^2 - 6x + 9 = (x - 3)^2 = 0 \implies x = 3 (invalid/extraneous). If k=−6k = -6, x2+6x+9=(x+3)2=0  ⟹  x=−3x^2 + 6x + 9 = (x + 3)^2 = 0 \implies x = -3 (valid).
The value k=6k = 6 makes x=3x = 3 the only solution, which is excluded. Therefore, only k=−6k = -6 yields exactly one valid real solution (x=−3x = -3).

Key Concept

Solving rational equations that simplify to quadratic equations, using the discriminant to find the number of solutions, and identifying extraneous solutions due to domain restrictions.
Estimated Time:2m 0s
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Topics

Quadratic Equations

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