Answer:
Two: [tex]\sqrt{d}[/tex] and [tex]\left(-\sqrt{d}\right)[/tex].
Step-by-step explanation:
The question is asking for the number of real numbers [tex]x[/tex] whose square is equal to [tex]d[/tex] (where [tex]d > 0[/tex].)
Naturally, the square root of [tex]d[/tex]: [tex]\sqrt{d}[/tex] would satisfy the requirement. [tex]\left(\sqrt{d}\right)^{2} = d[/tex].
However, keep in mind that [tex](x)^2[/tex] and [tex](-x)^{2}[/tex] have the same value. Therefore, if [tex]x =\sqrt{d}[/tex] would satisfy the equation, the opposite [tex]x = -\sqrt{d}[/tex] would also do.
Since [tex]d > 0[/tex], [tex]\sqrt{d}[/tex] and [tex]\left(-\sqrt{d}\right)[/tex] would both take real values.
Because [tex]d \ne 0[/tex], the value of [tex]\sqrt{d}[/tex] and [tex]\left(-\sqrt{d}\right)[/tex] would be distinct.
Hence, [tex]\sqrt{d}[/tex] and [tex]\left(-\sqrt{d}\right)[/tex] would correspond to the two real solutions of this equation.
