Find $\;$ $b$ $\;$ in the equation $\;$ $5x^2 + bx - 28 = 0$ $\;$ if the roots $x_1$ and $x_2$ of the equation are related as $5x_1 + 2x_2 = 1$ and $b$ is an integer.
Given quadratic equation: $\;\;\;$ $5x^2 + bx - 28 = 0$
Relation between the roots $x_1$ and $x_2$ of the given quadratic equation is
$5x_1 + 2x_2 = 1$ $\;\;\; \cdots \; (1)$
Sum of roots of the given quadratic equation: $\;$ $x_1 + x_2 = \dfrac{-b}{5}$ $\;\;\; \cdots \; (2)$
i.e. $\;$ $5x_1 + 5x_2 = \dfrac{-b}{5}$ $\;\;\; \cdots \; (2a)$
Solving equations $(1)$ and $(2a)$ simultaneously gives
$3x_2 = -b - 1$ $\implies$ $x_2 = \dfrac{-\left(b+1\right)}{3}$
Substituting the value of $x_2$ in equation $(2)$ gives
$x_1 = \dfrac{-b}{5} - x_2 = \dfrac{-b}{5} + \dfrac{b+1}{3}$ $\implies$ $x_1 = \dfrac{2b + 5}{15}$
Product of roots of the given quadratic equation: $\;$ $x_1 \cdot x_2 = \dfrac{-28}{5}$ $\;\;\; \cdots \; (3)$
Substituting the values of $x_1$ and $x_2$ in equation $(3)$ gives
$- \left(\dfrac{2b + 5}{15}\right) \left(\dfrac{b+1}{3}\right) = \dfrac{-28}{5}$
i.e. $\;$ $2b^2 + 7b + 5 = 252$
i.e. $\;$ $2b^2 + 7b - 247 = 0$
i.e. $\;$ $\left(b + 13\right) \left(2b - 19\right) = 0$
i.e. $\;$ $b = -13$ $\;$ or $\;$ $b = \dfrac{19}{2}$
$\because \;$ $b$ is an integer, $b = -13$ is the only acceptable solution.