Question

drag and drop the solutions given below into their correct places.

\\x^2 + 5x + 6 = 0\\
\\x^2 - x - 6 = 0\\
\\x^2 + x - 6 = 0\\
\\x^2 - 5x + 6 = 0\\

options:
\\x = 2, -3\\
\\x = 3, -2\\
\\x = 3, 2\\
\\x = -2, -3\\

Explanation:

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<reasoning>

Solve the first quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 + 5x + 6 = 0 \\ &(x + 2)(x + 3) = 0 \\ &x = -2, -3 \end{aligned}$$

\]

Solve the second quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 - x - 6 = 0 \\ &(x - 3)(x + 2) = 0 \\ &x = 3, -2 \end{aligned}$$

\]

Solve the third quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 + x - 6 = 0 \\ &(x + 3)(x - 2) = 0 \\ &x = 2, -3 \end{aligned}$$

\]

Solve the fourth quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 - 5x + 6 = 0 \\ &(x - 2)(x - 3) = 0 \\ &x = 3, 2 \end{aligned}$$

\]
</reasoning>

<answer>

No.	Problem	Answer
2	\(x^2 - x - 6 = 0\)	\(x = 3, -2\)
3	\(x^2 + x - 6 = 0\)	\(x = 2, -3\)
4	\(x^2 - 5x + 6 = 0\)	\(x = 3, 2\)

</answer>

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Answer:

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<reasoning>

Solve the first quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 + 5x + 6 = 0 \\ &(x + 2)(x + 3) = 0 \\ &x = -2, -3 \end{aligned}$$

\]

Solve the second quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 - x - 6 = 0 \\ &(x - 3)(x + 2) = 0 \\ &x = 3, -2 \end{aligned}$$

\]

Solve the third quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 + x - 6 = 0 \\ &(x + 3)(x - 2) = 0 \\ &x = 2, -3 \end{aligned}$$

\]

Solve the fourth quadratic equation

Using the Factoring Quadratic Trinomials and Quadratic Equations knowledge points
\[

$$\begin{aligned} &x^2 - 5x + 6 = 0 \\ &(x - 2)(x - 3) = 0 \\ &x = 3, 2 \end{aligned}$$

\]
</reasoning>

<answer>

No.	Problem	Answer
2	\(x^2 - x - 6 = 0\)	\(x = 3, -2\)
3	\(x^2 + x - 6 = 0\)	\(x = 2, -3\)
4	\(x^2 - 5x + 6 = 0\)	\(x = 3, 2\)

</answer>

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