Question

Equation x² + 5x - 7 = 0 has roots a and b. Equation 2x² + px + q = 0 has roots
a+land b + 1. Find p + q?

Answer 1

        $\text{\huge \bf \underline{Answer:}}$

        $\text{It is given that }\alpha \text{ and }\beta \text{ are the roots of the equation }x^{2} + 5x - 7 = 0$

        $\text{Applying the relationship between roots and coefficients}$

$\implies \: \text{Sum of roots} \longrightarrow \alpha + \beta = \dfrac{-b}{a} \longrightarrow \alpha + \beta = \dfrac{-5}{1} \longrightarrow \alpha + \beta = -5$

$\implies \: \text{Product of roots} \longrightarrow \alpha\beta = \dfrac{c}{a} \longrightarrow \alpha\beta = \dfrac{-7}{1} \longrightarrow \alpha\beta = -7$

        $\text{It is also given that }(\alpha + 1) \text{ and }(\beta + 1) \text{ are the roots of }2x^{2} + px + q = 0$

        $\text{Applying the relationship between roots and coefficients}$

$\implies \: \text{Sum of roots} \longrightarrow \alpha + 1+ \beta + 1 = \dfrac{-b}{a}$

$\implies \: \alpha +\beta + 2 = \dfrac{-p}{2}$

        $\text{We know that }\alpha + \beta = -5$

$\implies \: -5 + 2 = \dfrac{-p}{2}$

$\implies \:-3 = \dfrac{-p}{2}$

$\implies \:-3 \times 2 = -p$

$\implies \: -6 = -p$

$\implies \: p = 6$

$\implies \: \text{Product of roots} \longrightarrow (\alpha+1)(\beta + 1) =\dfrac{c}{a}$

        $\text{We know that }\alpha + \beta = -5 \text{ and }\alpha\beta = -7$

$\implies \: \alpha\beta + \alpha + \beta+ 1=\dfrac{q}{2}$

$\implies \: (-7) + (-5)+ 1=\dfrac{q}{2}$

$\implies \: -7 -5+ 1=\dfrac{q}{2}$

$\implies \: -11=\dfrac{q}{2}$

$\implies \: (-11) \times 2 = q$

$\implies \: q = -22$

$\implies \: \boxed{\boxed{p + q = 6 + (-22) = 6 - 22 = -16}}$