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Difficulty: MediumQuadratic Equations

In the quadratic equation x2−px+q=0x^2 - px + q = 0, pp and qq are positive constants. If the equation has exactly one real solution, what is the value of p2q\frac{p^2}{q}?

  1. A
    −4-4
  2. 44Answer
  3. C
    22
  4. D
    1616

Answer

The correct answer is 44.
The correct answer is 44. For any quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0 to have exactly one real solution, its discriminant Δ=b2−4ac\Delta = b^2 - 4ac must equal 00. In the equation x2−px+q=0x^2 - px + q = 0, we identify a=1a = 1, b=−pb = -p, and c=qc = q. Substituting these values into the discriminant formula gives (−p)2−4(1)(q)=0(-p)^2 - 4(1)(q) = 0, which simplifies to p2−4q=0p^2 - 4q = 0. Rearranging this equation gives p2=4qp^2 = 4q. Since qq is a positive constant, we can divide both sides of the equation by qq to isolate the ratio, yielding p2q=4\frac{p^2}{q} = 4.

Step-by-Step Solution

1
Recall the condition for a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0 to have exactly one real solution.
The discriminant, Δ=b2−4ac\Delta = b^2 - 4ac, must be equal to 00.
A quadratic equation has exactly one real solution if and only if its discriminant is zero, meaning the parabola is tangent to the x-axis.
2
Identify the coefficients of the given quadratic equation x2−px+q=0x^2 - px + q = 0 and substitute them into the discriminant formula.
Here, a=1a = 1, b=−pb = -p, and c=qc = q. Thus, Δ=(−p)2−4(1)(q)=p2−4q=0\Delta = (-p)^2 - 4(1)(q) = p^2 - 4q = 0.
To apply the one-solution condition specifically to the coefficients of the given equation.
3
Solve the equation p2−4q=0p^2 - 4q = 0 for the ratio p2q\frac{p^2}{q}.
p2=4q  ⟹  p2q=4p^2 = 4q \implies \frac{p^2}{q} = 4.
Isolate the target expression by dividing both sides of the equation by qq, which is valid since qq is a positive constant and thus non-zero.

Key Concept

Evaluating the discriminant of a quadratic equation to determine the number of real solutions.
Estimated Time:1m 30s
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Topics

Quadratic Equations

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