Match the value of the discriminant with the number of solutions Value of Discriminant Number of Solutions \begin{tabular}{ll}$$ -\vee 12 $$ & a. No Real Solution \ $$ \begin{array}{ll}-\vee & ext { b. } 2 ext { Real Solutions }\end{array} $$ \ \hline\end{tabular}

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Match the value of the discriminant with the number of solutions Value of Discriminant Number of Solutions \begin{tabular}{ll}$$ -\vee 12 $$ & a. No Real Solution \ $$ \begin{array}{ll}-\vee & 	ext { b. } 2 	ext { Real Solutions }\end{array} $$ \ \hline\end{tabular}

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To match the value of the discriminant with the number of solutions, we need to understand how the discriminant ($$D$$) of a quadratic equation relates to the nature of its solutions. The discriminant is given by the formula: $$ D = b^2 - 4ac $$ where $$a$$, $$b$$, and $$c$$ are the coefficients of the quadratic equation $$ax^2 + bx + c = 0$$. The relationship between the value of the discriminant and the number of solutions is as follows: 1. If $$D < 0$$ (which includes values like $$-12$$), there are **no real solutions**. The solutions are complex. 2. If $$D = 0$$, there is **exactly one real solution** (a repeated root). 3. If $$D > 0$$, there are **two distinct real solutions**. Now, let's match the given values: - For $$D < 0$$ (e.g., $$D = -12$$), the number of solutions is **a. No Real Solution**. - For $$D = 0$$, the number of solutions would be **exactly one real solution** (not listed in your table). - For $$D > 0$$, the number of solutions is **b. 2 Real Solutions**. Thus, the matches are: - $$D < 0$$ (e.g., $$-12$$) corresponds to **a. No Real Solution**. - $$D > 0$$ corresponds to **b. 2 Real Solutions**. In summary: - $$D < 0$$ → a. No Real Solution - $$D > 0$$ → b. 2 Real Solutions - $$D < 0$$ (e.g., $$-12$$) corresponds to **a. No Real Solution**. - $$D > 0$$ corresponds to **b. 2 Real Solutions**.

Match the value of the discriminant with the number of solutions Value of Discriminant Number of Solutions \begin{tabular}{ll}\( -\vee 12 \) & a. No Real Solution \\ \( \begin{array}{ll}-\vee & \text { b. } 2 \text { Real Solutions }\end{array} \) \\ \hline\end{tabular}

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