Chapter 12.5, Problem 8WE

To determine

Solve the quadratic equation.

Answer to Problem 8WE

  $x\in \left\{-3,-\frac{1}{2}\right\}$

Explanation of Solution

Given:

The equation is :

  $2x^2+7x+3=0$

Calculation:

The given equation $2x^2+7x+3=0$ is of the type $a{x}^2+bx+c=0$ , where a = 2 and b = 7, c = 3 .

  $b^2-4ac={(7)}^2-4(2)(3)=49-24=25$ , which is perfect square .

So, it can be solved by using the Factoring the middle term.

  $\begin{array}{l}2x^2+7x+3=0\\ \Rightarrow 2x^2+6x+x+3=0\\ \Rightarrow 2x(x+3)+1(x+3)=0\\ \Rightarrow (2x+1)(x+3)=0\\ \Rightarrow 2x+1=0\text{ and }x+3=0\\ \Rightarrow x=-\frac{1}{2}\text{ and }x=-3\end{array}$

So, $x\in \left\{-3,-\frac{1}{2}\right\}$