EXERCISE 10 (a) Discuss the nature of the roots of the following quadratic equations, without solving the equations: \( \begin{array}{llll}\text { (1) } x^{2}=0 & \text { (2) } x^{2}=16 & \text { (3) } x^{2}=7 \\ \text { (4) } x^{2}=-4 & \text { (5) }(x-2)^{2}=4 & \text { (6) }(x+3)^{2}=12 \\ \text { (7) } \quad(x+1)^{2}=0 & \text { (8) }(x-5)^{2}=-1 & \text { (9) } x^{2}-9=0\end{array} \)
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In the equations provided, we can describe the nature of the roots based on the relationships involving squares and constants. For \(x^{2}=0\) (1), there is one real root at \(x=0\) as the square of a number can only be zero if the number itself is zero. For \(x^{2}=16\) (2), there are two distinct real roots: \(4\) and \(-4\) since both squares yield \(16\). In the third equation \(x^{2}=7\) (3), again we find two distinct real roots: \(\sqrt{7}\) and \(-\sqrt{7}\). In contrast, in \(x^{2}=-4\) (4), there are no real roots because a square cannot be negative, indicating complex numbers are involved. For \((x-2)^{2}=4\) (5) and \((x+3)^{2}=12\) (6), both yield two distinct real roots, indicating the roots are found at \(x=2\pm2\) and \(x=-3\pm\sqrt{12}\), respectively. Equation \((x+1)^{2}=0\) (7) has a double root at \(x=-1\) because it is a perfect square. Lastly, \((x-5)^{2}=-1\) (8) shows no real roots, indicating that solutions would require complex numbers, while \(x^{2}-9=0\) (9) gives us two distinct real roots \(3\) and \(-3\).