Which quadratic expression is positive for all real values of x?

1. INFER the key mathematical condition

For a quadratic \(\mathrm{ax^2 + bx + c}\) to be positive for all real values of \(\mathrm{x}\), we need two things:

• The parabola must open upward: \(\mathrm{a \gt 0}\)
• The parabola must never touch or cross the x-axis: discriminant \(\mathrm{b^2 - 4ac \lt 0}\)

Why both conditions? If \(\mathrm{a \lt 0}\), the parabola opens downward and will eventually become negative. If the discriminant \(\mathrm{\geq 0}\), the parabola touches or crosses the x-axis, meaning it equals zero or becomes negative somewhere.

2. SIMPLIFY by checking each option systematically

Let's test each choice against both conditions:

Option (A): \(\mathrm{x^2 + 12x - 36}\)

• \(\mathrm{a = 1 \gt 0}\) ✓ (opens upward)
• Discriminant \(\mathrm{= 12^2 - 4(1)(-36)}\)
  \(\mathrm{= 144 + 144}\)
  \(\mathrm{= 288 \gt 0}\) ✗

This has real roots, so it changes sign.

Option (B): \(\mathrm{x^2 - 12x + 36}\)

• \(\mathrm{a = 1 \gt 0}\) ✓ (opens upward)
• Discriminant \(\mathrm{= (-12)^2 - 4(1)(36)}\)
  \(\mathrm{= 144 - 144}\)
  \(\mathrm{= 0}\) ✗

This touches the x-axis at exactly one point, so it equals zero there.

Option (C): \(\mathrm{-3x^2 - 12x + 36}\)

• \(\mathrm{a = -3 \lt 0}\) ✗ (opens downward)

Since the parabola opens downward, it cannot be positive for all real \(\mathrm{x}\).

Option (D): \(\mathrm{3x^2 - 12x + 36}\)

• \(\mathrm{a = 3 \gt 0}\) ✓ (opens upward)
• Discriminant \(\mathrm{= (-12)^2 - 4(3)(36)}\)
  \(\mathrm{= 144 - 432}\)
  \(\mathrm{= -288 \lt 0}\) ✓

Both conditions satisfied!

Answer: D

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students often check only one condition, typically just looking at whether \(\mathrm{a \gt 0}\), without considering the discriminant. They might think 'if the parabola opens upward, it's always positive' and select choice (A) or (B).

This incomplete reasoning leads them to select Choice (A) (\(\mathrm{x^2 + 12x - 36}\)) or Choice (B) (\(\mathrm{x^2 - 12x + 36}\)) without realizing these expressions can equal zero or become negative.

Second Most Common Error:

Conceptual confusion about discriminant = 0: Students may know they need to check the discriminant but forget that when it equals zero, the quadratic still touches the x-axis (equals zero at one point). They incorrectly think 'no negative discriminant means always positive.'

This leads them to select Choice (B) (\(\mathrm{x^2 - 12x + 36}\)) because they see \(\mathrm{a \gt 0}\) and discriminant \(\mathrm{= 0}\), not realizing that 'equals zero somewhere' violates 'positive for ALL real \(\mathrm{x}\).'

The Bottom Line:

This problem requires understanding that 'positive for all real \(\mathrm{x}\)' is very restrictive - the quadratic can never equal zero or be negative anywhere, which means it must hover completely above the x-axis.