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$\pm \sqrt{{\color{#0000FF}{4}}} =\pm 2$ | |

$2^2 = 2(2) = {\color{#0000FF}{4}}$ | $(\text{-}2)^2 = \text{-}2(\text{-}2) = {\color{#0000FF}{4}}$ |

With this in mind, we can write the required equation. $\begin{gathered} x^2 = 4 \end{gathered}$ Note that this is just an example. Since any $d>0$ will satisfy the requirements of the exercise, there are infinitely many solutions.

b

In Part A we found that $x=\pm \sqrt{d}.$ If we want our equation to have just one real solution, then $d$ should be $0.$ This is because $0$ is the only real number that doesn't have an opposite to allow for two solutions. $\begin{gathered} x^2 = 0 \end{gathered}$

c

If we want the equation to have no real solutions, we can use any negative number for $d.$ This is because there is not a real number such that when multiplied by itself results in a negative number. With this in mind, we can write the required equation. $\begin{gathered} x^2 = \text{-}4 \end{gathered}$ Notice that this is just an example. Since any negative number will satisfy the requirements of the exercise, there are infinitely many solutions.