Chapter 4.2, Problem 11CYU

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 11CYU

  $x_1=\left(-5,-4\right)\text{ and }{x}_2=\left(5,6\right)$ .

Explanation of Solution

Given:

The given equation is $x^2-20=2+x$ .

Calculation:

The given equation is expressed in to standard form:

  $\begin{array}{l}{x}^2-20=2+x\{\text{given equation}}\\ x^2-x-22=0\left\{\text{equate one side to 0}\right\}\end{array}$ .

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

  

Looking at the graph, the parabola crosses the x -axis, therefore, the equation has the solution.

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

  $\begin{array}{l}a{x}^2+bx+c=0\\ \therefore a=1,b=-1,c=-22\\ x_v=\frac{-b}{2a}=\frac{1}{2\left(1\right)}=0.5\end{array}$

$x$	$y=x^2-x-22$
$-6$	$y={\left(-6\right)}^2-\left(-6\right)-22=20$
$-5$	$y={\left(-5\right)}^2-\left(-5\right)-22=8$
$-4$	$y={\left(-4\right)}^2-\left(-4\right)-22=-2$
$-3$	$y={\left(-3\right)}^2-\left(-3\right)-22=-10$
$-2$	$y={\left(-2\right)}^2-\left(-2\right)-22=-16$
$-1$	$y={\left(-1\right)}^2-\left(-1\right)-22=-20$
$0$	$y={\left(0\right)}^2-\left(0\right)-22=-22$
$0.5$	$y={\left(0.5\right)}^2-\left(0.5\right)-22=-22.25$
$1$	$y={\left(1\right)}^2-\left(1\right)-22=-22$
$2$	$y={\left(2\right)}^2-\left(2\right)-22=-20$
$3$	$y={\left(3\right)}^2-\left(3\right)-22=-16$
$4$	$y={\left(4\right)}^2-\left(4\right)-22=-10$
$5$	$y={\left(5\right)}^2-\left(5\right)-22=-2$
$6$	$y={\left(6\right)}^2-\left(6\right)-22=8$

It is noticed that the equation has two real roots.

  $\begin{array}{l}{x}_1=\left(-5,-4\right)\\ x_2=\left(5,6\right)\end{array}$

Hence, the given equation has two real roots. i.e. $x_1=\left(-5,-4\right)\text{ and }{x}_2=\left(5,6\right)$ .