Chapter L, Problem 1DE

Solve by graphing: $x^2+2x-3>0.$

To determine

To calculate: The solution of the inequality $x^2+2x-3>0$ by graphing.

Answer to Problem 1DE

Solution:

The solution of the inequality $x^2+2x-3>0$ is $\left\{x|-3>x\text{ or }x>1\right\}$ or $\left(-\infty ,-3\right)\cup \left(1,\infty \right)$.

Explanation of Solution

Given Information:

The provided inequality is $x^2+2x-3>0$.

Formula used:

Steps to solve the inequality by graphing.

1. Replace the inequality sign by equal sign.

2. Find the leading coefficient in the inequality.

If the leading coefficient in the inequality is negative, then the graph opens down and if the leading coefficient is positive, then the graph opens up.

3. Set the polynomial equal to zero to find the x-intercept and also find y-intercept.

4. Plot the graph of the equation.

Zero product rule is such that if $a\cdot b=0$, then either $a=0$ or $b=0$.

Calculation:

Consider the inequality, $x^2+2x-3>0$

Now, the inequality $x^2+2x-3>0$ is represented as $x^2+2x-3=0$.

So, the equation becomes $y=x^2+2x-3$.

Put the equation equals to zero to find the x-intercepts,

$\begin{array}{c}{x}^2+2x-3=0\\ x^2+3x-x-3=0\\ x\left(x+3\right)-1\left(x+3\right)=0\\ \left(x+3\right)\left(x-1\right)=0\end{array}$

Use the zero product rule and simplify the above equation as,

$\begin{array}{c}x+3=0\text{ or }x-1=0\\ x=-3\text{ or }x=1\end{array}$

The x-intercepts are $\left(-3,0\right)$ and $\left(1,0\right)$.

The leading coefficient in the polynomial $x^2+2x-3$ is 1 which is positive due to which the graph opens upward.

Now, find the point where the graph intersects y-axis.

Substitute $x=0$ in the equation $y=x^2+2x-3$,

$\begin{array}{c}y={\left(0\right)}^2+2\left(0\right)-3\\ =-3\end{array}$

So, the y-intercept is $\left(0,-3\right)$.

The table provided below shows the values of y for different values of x,

$\begin{array}{cc}x& y\\ 0& -3\\ -3& 0\\ 1& 0\end{array}$

The graph of the equation $y=x^2+2x-3$ is shown below:

From the observation of the graph, the values of y is positive when $x$ is less than $-3$ or when $x$ is greater than 1.

So, the values for which y be positive are $x<-3\text{ or }x>1$.

Therefore, the solution of the inequality $x^2+2x-3>0$ is $\left\{x|-3>x\text{ or }x>1\right\}$ or $\left(-\infty ,-3\right)\cup \left(1,\infty \right)$.