The quadratic expression $(2X+1)^2-pX+q\neq 0$ for any real $X$, if
- $p^2-16p-8q < 0$
- $p^2-8p+16q < 0$
- $p^2-8p-16q < 0$
- $p^2-16p+8q < 0$
Reveal answer
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C
Correct answer
Explanation
For the expression to never be zero, the discriminant of the quadratic (2X+1)^2 - pX + q = 0 must be negative. 4X^2 + 4X + 1 - pX + q = 4X^2 + (4-p)X + (1+q) = 0. D = (4-p)^2 - 4 * 4 * (1+q) < 0. 16 - 8p + p^2 - 16 - 16q < 0 => p^2 - 8p - 16q < 0.
AI explanation
Expanding the expression yields 4x^2 + (4 - p)x + (1 + q) = 0. For this quadratic expression to never equal zero for any real x, its discriminant must be strictly less than zero, so we evaluate D < 0 using D = b^2 - 4ac. Substituting the coefficients gives (4 - p)^2 - 4(4)(1 + q) < 0, which expands to 16 - 8p + p^2 - 16 - 16q < 0. Simplifying this inequality results in p^2 - 8p - 16q < 0.