To find the values of \( p \) and \( q \) for the quadratic equation \( 2x^2 + (p + 1)x + q = 0 \) given that the roots are \( 1 \) and \( 3 \), we can use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots.
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Sum of the roots: According to Vieta's formulas, the sum of the roots \( r_1 + r_2 \) (where \( r_1 = 1 \) and \( r_2 = 3 \)) is given by: \[ r_1 + r_2 = -\frac{b}{a} = -\frac{(p + 1)}{2} \] Therefore, \[ 1 + 3 = -\frac{(p + 1)}{2} \] Simplifying this, we have: \[ 4 = -\frac{(p + 1)}{2} \] Multiplying both sides by \(-2\): \[ -8 = p + 1 \] Thus, \[ p = -9. \]
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Product of the roots: The product of the roots \( r_1 \cdot r_2 \) is given by: \[ r_1 \cdot r_2 = \frac{c}{a} = \frac{q}{2} \] Therefore, \[ 1 \cdot 3 = \frac{q}{2} \] Which simplifies to: \[ 3 = \frac{q}{2} \] Multiplying both sides by \( 2 \): \[ q = 6. \]
So, the values of \( p \) and \( q \) are: \[ p = -9 \quad \text{and} \quad q = 6. \]
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