Chapter 4.2, Problem 41PPS

To determine

To solve: the given equation. If exact roots cannot be found, state the consecutive integers between which the roots are located.

Answer to Problem 41PPS

  $x=2.5,x=-3$ .

Explanation of Solution

Given:

The given equation is $2x^2+x=15$ .

Calculation:

The given equation is $2x^2+x-15=0$ .

Now, to find the zeros of the given equation by using determinant method. Therefore,

  $\begin{array}{l}X=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\left\{\text{determinant formula}\right\}\\ X=\frac{-1\pm \sqrt{{\left(-1\right)}^2-4\left(2\right)\left(-15\right)}}{2\left(2\right)}\{\text{substitues the value}}\\ X=\frac{-1\pm \sqrt{1-\left(-120\right)}}{4}\{\text{simplify}}\\ X=\frac{-1\pm \sqrt{121}}{4}=\frac{-1\pm 11}{4}\end{array}$

Therefore, either $x_1=\frac{-1+11}{4}$ or $x_2=\frac{-1-11}{4}$

When $x_1=\frac{-1+11}{4}$ , then $x=2.5$ .

When $x_2=\frac{-1-11}{4}$ , then $x=-3$ .

Thus, the zeros of the function are $x=2.5,x=-3$ .

Hence, the solution of the given equation is $x=2.5,x=-3$ .