Question

solve for x and graph the solution.
5x² + 30x + 25 > 0
plot the endpoints. select an endpoint to change it from closed to open. select the middle of a segment, ray, or line to delete it.

Explanation:

Step1: Simplify the inequality

Divide the inequality \(5x^{2}+30x + 25>0\) by \(5\) to get \(x^{2}+6x + 5>0\).
Factor the quadratic: \(x^{2}+6x + 5=(x + 1)(x + 5)>0\).

Step2: Find the roots

Set \((x + 1)(x + 5)=0\). Using the zero - product property \(x+1 = 0\) gives \(x=-1\), and \(x + 5=0\) gives \(x=-5\).

Step3: Test intervals

We have three intervals to test: \((-\infty,-5)\), \((-5,-1)\), and \((-1,\infty)\).

• For \(x=-6\) (in the interval \((-\infty,-5)\)): \((-6 + 1)(-6 + 5)=(-5)\times(-1)=5>0\).
• For \(x=-3\) (in the interval \((-5,-1)\)): \((-3 + 1)(-3 + 5)=(-2)\times2=-4<0\).
• For \(x=0\) (in the interval \((-1,\infty)\)): \((0 + 1)(0 + 5)=1\times5 = 5>0\).

Answer:

The solution of the inequality \(5x^{2}+30x + 25>0\) is \(x<-5\) or \(x>-1\). On the number - line, we have open circles at \(x=-5\) and \(x=-1\). The graph consists of two rays: one extending to the left from \(x = - 5\) (not including \(x=-5\)) and one extending to the right from \(x=-1\) (not including \(x=-1\)).