Question

5
$(x + 81)^2 = -256$
how many distinct real solutions does the given equation have ?
a exactly one
b exactly two
c infinitely many
d zero

Explanation:

Step1: Recall the property of squares

The square of any real number \(a\) (i.e., \(a^{2}\)) is always non - negative, that is \(a^{2}\geq0\) for all real numbers \(a\). In the given equation \((x + 81)^{2}=-256\), the left - hand side \((x + 81)^{2}\) represents the square of a real number (since \(x\) is a real number, \(x + 81\) is also a real number). The right - hand side is \(-256\), which is a negative number.

Step2: Analyze the equation for real solutions

Since the square of a real number cannot be negative, there is no real number \(x\) such that \((x + 81)^{2}=-256\) holds. So the number of distinct real solutions of the equation is zero.

Answer:

D. Zero