Question

In Problems 18–19, solve each quadratic inequality.

$3x^2\ge 14x+5$

Short answer

The solution of inequality is$\left\{\overline{)x}x\le -\frac{1}{3}\mathrm{or}x\ge 5\right\}$ or interval notation$(-\infty ,-\frac{1}{3}]\mathrm{or}[5,\infty )$.

Step-by-step solution

Step 1. Given information. 

Solve the given quadratic inequality.

$3x^2\ge 14x+5$

Step 2. Graph the function.

Rearrange the function.

$$\begin{gathered} 3x^2\ge 14x+5 \\ 3x^2-14x-5\ge 0 \end{gathered}$$

Graph the function $f\left(x\right)=3x^2-14x-5$

Step 3. Find intercept. 

y- intercept: $f\left(0\right)=3{\left(0\right)}^2-14\left(0\right)-5=-5$

x- intercept: Solve$f\left(x\right)=0$

$$\begin{gathered} 3x^2-14x-5=0 \\ 3x^2-15x+x-5=0 \\ 3x(x-5)+1(x-5)=0 \\ (x-5)(3x+1)=0 \\ x=5,-\frac{1}{3} \end{gathered}$$

So, the x-intercept are $x_1=5\mathrm{and}{x}_2=-\frac{1}{3}$

The solution set is $\left\{\overline{)x}x\le -\frac{1}{3}\mathrm{or}x\ge 5\right\}$ and in interval notation$(-\infty ,-\frac{1}{3}]\mathrm{or}[5,\infty )$.

Step 4. Conclusion. 

The solution set of inequality is $\left\{\overline{)x}x\le -\frac{1}{3}\mathrm{or}x\ge 5\right\}$or in interval notation $(-\infty ,-\frac{1}{3}]\mathrm{or}[5,\infty )$.