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The question asks for the values of \( p \) and \( q \) in the quadratic equation:

\[
x^2 + px + q = 0
\]

It provides two roots of the equation: \( -2 \) and \( 3 \).

Using Vieta's formulas for a quadratic equation \( x^2 + px + q = 0 \), where the roots are \( r_1 \) and \( r_2 \):

1. The sum of the roots is equal to \( -p \):  
   \[
   r_1 + r_2 = -p
   \]
   In this case:
   \[
   -2 + 3 = 1 \implies p = -1
   \]

2. The product of the roots is equal to \( q \):  
   \[
   r_1 \times r_2 = q
   \]
   In this case:
   \[
   -2 \times 3 = -6 \implies q = -6
   \]

Therefore, the values of \( p \) and \( q \) are:

\[
p = -1, \quad q = -6
\]

Would you like further explanation on how Vieta's formulas work?

Here are 5 questions to expand your understanding:
1. How do Vieta’s formulas apply to higher-degree polynomials?
2. What are the conditions for a quadratic equation to have real roots?
3. How can you derive the quadratic formula from the standard form of a quadratic equation?
4. Can a quadratic equation have complex roots, and how would that affect \( p \) and \( q \)?
5. How do you factor a quadratic equation based on its roots?

**Tip:** The discriminant \( \Delta = b^2 - 4ac \) helps determine the nature of the roots—real or complex—of a quadratic equation.

若方程x²+px+q=0的两个根是-2和3，则p,q的值分别为______。

Learn how to solve for p and q in the quadratic equation x² + px + q = 0 where the roots are -2 and 3. Using Vieta's formulas, we calculate that p = -1 and q = -6. Suitable for students in Grades 7-9.

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