Question

solve for x and graph the solution.
2x² - 12x + 10 ≤ 0
plot the endpoints. select an endpoint to change it from closed to open. select a segment, ray, or line to delete it.

Explanation:

Step1: Simplify the inequality

Divide the entire inequality \(2x^{2}-12x + 10\leq0\) by \(2\) to get \(x^{2}-6x + 5\leq0\).
Factor the quadratic expression: \(x^{2}-6x + 5=(x - 1)(x - 5)\). So the inequality becomes \((x - 1)(x - 5)\leq0\).

Step2: Find the critical points

Set \((x - 1)(x - 5)=0\). Using the zero - product property \(a\times b = 0\) implies \(a = 0\) or \(b = 0\).
If \(x-1=0\), then \(x = 1\); if \(x - 5=0\), then \(x = 5\).

Step3: Test intervals

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

• For the interval \((-\infty,1)\), let \(x=0\). Then \((0 - 1)(0 - 5)=( - 1)\times(-5)=5>0\).
• For the interval \((1,5)\), let \(x = 3\). Then \((3 - 1)(3 - 5)=(2)\times(-2)=-4\leq0\).
• For the interval \((5,\infty)\), let \(x=6\). Then \((6 - 1)(6 - 5)=(5)\times(1)=5>0\).

Answer:

The solution of the inequality \(2x^{2}-12x + 10\leq0\) is \(1\leq x\leq5\). On the number - line, we plot closed circles at \(x = 1\) and \(x = 5\) (since the inequality includes equality) and shade the region between them.