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Difficulty: MediumQuadratic Equations and the Quadratic Formula

If kk is a non-zero constant, for what value of kk does the quadratic equation (x−3)2=kx(x - 3)^2 = kx have exactly one real solution?

  1. A
    66
  2. B
    −3-3
  3. −12-12Answer
  4. D
    −6-6
  5. E
    1212

Answer

−12-12
Expanding (x−3)2(x - 3)^2 yields x2−6x+9=kxx^2 - 6x + 9 = kx. Subtracting kxkx from both sides and grouping like terms gives x2−(6+k)x+9=0x^2 - (6 + k)x + 9 = 0. For a quadratic equation to have exactly one real solution, its discriminant must be equal to zero. Thus, we set (−(6+k))2−4(1)(9)=0(-(6+k))^2 - 4(1)(9) = 0, which simplifies to (6+k)2−36=0(6+k)^2 - 36 = 0. Taking the square root of both sides gives 6+k=66 + k = 6 or 6+k=−66 + k = -6. Solving these equations gives k=0k = 0 or k=−12k = -12. Since the problem specifies that kk is a non-zero constant, the correct value is −12-12.

Step-by-Step Solution

1
Expand the squared binomial on the left side of the equation.
x2−6x+9=kxx^2 - 6x + 9 = kx
Expanding the binomial (x−3)2(x - 3)^2 allows us to rewrite the equation in a form where we can group terms.
2
Move kxkx to the left side and group the xx terms to write the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−(6+k)x+9=0x^2 - (6 + k)x + 9 = 0, where a=1a = 1, b=−(6+k)b = -(6 + k), and c=9c = 9.
The coefficients aa, bb, and cc must be identified from the standard quadratic form to calculate the discriminant.
3
Set the discriminant b2−4acb^2 - 4ac equal to 00 and solve for kk.
(−(6+k))2−4(1)(9)=0  ⟹  (6+k)2−36=0  ⟹  (6+k)2=36  ⟹  6+k=±6(-(6+k))^2 - 4(1)(9) = 0 \implies (6+k)^2 - 36 = 0 \implies (6+k)^2 = 36 \implies 6+k = \pm 6. This yields k=0k = 0 or k=−12k = -12. Since kk is non-zero, k=−12k = -12.
A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

Key Concept

Using the discriminant (b2−4ac=0b^2 - 4ac = 0) to determine when a quadratic equation has exactly one real solution.

Alternative Method

Alternatively, one can recognize that the equation (x−3)2=kx(x-3)^2 = kx can be written as x2−(6+k)x+9=0x^2 - (6+k)x + 9 = 0. For a quadratic equation with a leading coefficient of 11 and a constant term of 99 to have exactly one real solution, it must be a perfect square trinomial. A perfect square trinomial of the form x2+bx+9x^2 + bx + 9 must have b=±6b = \pm 6. Setting the middle coefficient equal to these values gives −(6+k)=6  ⟹  k=−12-(6+k) = 6 \implies k = -12 or −(6+k)=−6  ⟹  k=0-(6+k) = -6 \implies k = 0. Since kk is non-zero, k=−12k = -12.
Estimated Time:1m 30s
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Quadratic Equations and the Quadratic Formula

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