Question

If two zeros are given as $\alpha \text{ and }\beta$ of the polynomial $2{{x}^{2}}+3x+5=0$ then find the value of $\dfrac{\alpha +\beta }{\alpha \beta }$?

Answer 1

Hint: For solving this problem, first we find the roots of the equation by using the discriminant method. Discriminant method is used because the roots are complex. Since we have the individual value of the roots, we can evaluate the required thing easily.
Complete step-by-step solution -
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If the quadratic function is set equal to zero, then the result is a quadratic equation. The solutions to the univariate equation are called the roots of the univariate function.
The roots of the equation by using the discriminant could be expressed as: $D=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
According to the problem statement, we are given in equation $2{{x}^{2}}+3x+5=0$.
Therefore, the roots of the equation can be expressed as: $D=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By putting a = 2, b = 3 and c = 5, we get
$\begin{aligned} & D=\dfrac{-3\pm \sqrt{{{3}^{2}}-4\left( 2 \right)\left( 5 \right)}}{2\times 2} \\\ & D=\dfrac{-3\pm \sqrt{9-40}}{4} \\\ & D=\dfrac{-3\pm i\sqrt{31}}{4} \\\ \end{aligned}$
The two roots of the equation are $\dfrac{-3+i\sqrt{31}}{4},\dfrac{-3-i\sqrt{31}}{4}$.
Now, the sum of the roots:
$\begin{aligned} & \dfrac{-3+i\sqrt{31}}{4}+\dfrac{-3-i\sqrt{31}}{4} \\\ & \Rightarrow \dfrac{-6}{4}=\dfrac{-3}{2} \\\ \end{aligned}$
The product of the roots:
$\begin{aligned} & \dfrac{-3+i\sqrt{31}}{4}\times \dfrac{-3-i\sqrt{31}}{4} \\\ & \Rightarrow \dfrac{\left( -3+i\sqrt{31} \right)\left( -3-i\sqrt{31} \right)}{16} \\\ \end{aligned}$
By using the identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, we get

& \Rightarrow \dfrac{{{\left( -3 \right)}^{2}}-{{\left( i\sqrt{31} \right)}^{2}}}{16} \\\ & \Rightarrow \dfrac{9+31}{16}=\dfrac{40}{16} \\\ & \Rightarrow \dfrac{5}{2} \\\ \end{aligned}$$ Finally putting in the required expression, we get $\begin{aligned} & \dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{\dfrac{-3}{2}}{\dfrac{5}{2}} \\\ & \Rightarrow \dfrac{-3}{5} \\\ \end{aligned}$ Hence, the value of is $\dfrac{-3}{5}$. Note: This problem can alternatively be solved by using the concept of sum of roots and product of roots as the sum of roots can be expressed as $\dfrac{-b}{a}$ and the product of roots can be expressed as $\dfrac{c}{a}$ from the general expression $a{{x}^{2}}+bx+c$. By using these values, we can directly evaluate the expression.