Chapter 4.2, Problem 10CYU

To determine

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

Answer to Problem 10CYU

  $x_1=\left(-2,-1\right)\text{ and }{x}_2=\left(-1,0\right)$ .

Explanation of Solution

Given:

The given equation is $9-x^2=12$ .

Calculation:

The given equation is expressed in to standard form:

  $\begin{array}{l}5x^2+10x-4=-6\{\text{given equation}}\\ 5x^2+10x+2=0\left\{\text{add 6 on both sides}\right\}\end{array}$ .

Now, graph the related quadratic function to show that x -intercept of the graph. Therefore,

  

Looking at the graph, the parabola did not cross the x -axis, therefore, the equation has the solution.

Identify $a,b,c$ and then find the vertex $V\left(x_v,y_v\right)$ .

  $\begin{array}{l}a{x}^2+bx+c=0\\ \therefore a=5,b=10,c=2\\ x_v=\frac{-b}{2a}=\frac{-10}{2\left(5\right)}=-1\end{array}$

$x$	$y=5x^2+10x+2$
$-3$	$y=5{\left(-3\right)}^2+10\left(-3\right)+2=17$
$-2$	$y=5{\left(-2\right)}^2+10\left(-2\right)+2=2$
$-1$	$y=5{\left(-1\right)}^2+10\left(-1\right)+2=-3$
$0$	$y=5{\left(0\right)}^2+10\left(0\right)+2=2$

It is noticed that the equation has two real roots.

  $\begin{array}{l}{x}_1=\left(-2,-1\right)\\ x_2=\left(-1,0\right)\end{array}$

Hence, the given equation has two real roots. i.e. $x_1=\left(-2,-1\right)\text{ and }{x}_2=\left(-1,0\right)$ .