Chapter 6.7, Problem 17PPS

To determine

Solve the given equation. State the number and type of roots.

Answer to Problem 17PPS

The equation $2x^2+x-6=0$ has two roots, both are real $-2$ and $\frac{3}{2}$ .

Explanation of Solution

Given information:

The equation $2x^2+x-6=0$ .

Formula used:

Splitting the middle term: To solve a quadratic equation $a{x}^2+bx+c=0$ , the middle term of equation is splitted in such a way that the product of coefficient of its parts is equal to product of coefficient of first and third term.

  $2x^2+x-6=0$

Recall that in splitting the middle term, the middle term of equation is splitted in such a way that the product of coefficient of its parts is equal to product of coefficient of first and third term.

  $\begin{array}{l}2x^2+4x-3x-6=0\\ \\ 2x\left(x+2\right)-3\left(x+2\right)=0\\ \\ \left(2x-3\right)\left(x+2\right)=0\end{array}$

By the above equation, we get that one or both terms $\left(2x-3\right)$ and $\left(x+2\right)$ are zero,

  $\begin{array}{l}2x-3=0\\ \\ 2x=3\\ \\ x=\frac{3}{2}\end{array}$

  $\begin{array}{l}x+2=0\\ \\ x=-2\end{array}$

The equation $2x^2+x-6=0$ has two roots, both are real $-2$ and $\frac{3}{2}$ .