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Difficulty: MediumQuadratic Equations and the Quadratic Formula

For the quadratic equation 1.5x2−5x+c=01.5x^2 - 5x + c = 0, where cc is a real constant, the equation has two non-real complex solutions. Which of the following inequalities represents all possible values of cc?

  1. c>256c > \frac{25}{6}Answer
  2. B
    c<256c < \frac{25}{6}
  3. C
    c>−256c > -\frac{25}{6}
  4. D
    c>503c > \frac{50}{3}
  5. E
    c<−256c < -\frac{25}{6}

Answer

c>256c > \frac{25}{6}
For a quadratic equation to have two non-real complex solutions, its discriminant must be negative. Substituting a=1.5a = 1.5, b=−5b = -5, and the constant cc into the discriminant formula b2−4ac<0b^2 - 4ac < 0 gives 25−6c<025 - 6c < 0. Solving this inequality results in the requirement that the constant must be strictly greater than twenty-five sixths.

Step-by-Step Solution

1
Identify the coefficients of the quadratic equation 1.5x2−5x+c=01.5x^2 - 5x + c = 0 in the standard form ax2+bx+c=0ax^2 + bx + c = 0.
a=1.5a = 1.5, b=−5b = -5, and c=cc = c.
These coefficients are required to compute the discriminant.
2
Set up the condition for the quadratic equation to have two non-real complex solutions using the discriminant Δ=b2−4ac\Delta = b^2 - 4ac.
The discriminant must be strictly negative: (−5)2−4(1.5)(c)<0(-5)^2 - 4(1.5)(c) < 0.
A quadratic equation has non-real complex solutions if and only if its discriminant is negative.
3
Simplify the inequality and solve for cc.
25−6c<0  ⟹  25<6c  ⟹  c>25625 - 6c < 0 \implies 25 < 6c \implies c > \frac{25}{6}.
Isolating cc yields the range of values that satisfy the condition.

Key Concept

A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has two non-real complex solutions if and only if its discriminant, Δ=b2−4ac\Delta = b^2 - 4ac, is strictly less than zero.
Estimated Time:1m 30s
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Topics

Quadratic Equations and the Quadratic Formula

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