Question

Solve each inequality algebraically.

$3(x^2-2)<2{(x-1)}^2+x^2$

Short answer

Solution set and the interval is:

$$\begin{gathered} \left\{x\right|x<2\left\} \\ \right(-\infty ,2) \end{gathered}$$

Step-by-step solution

Step 1. Change the inequality to equal to zero. 

Change the inequality equal to zero to make it easier and then get the value of x.

First arrange the inequality,

$\begin{array}{r}3\left(x^2-2\right)<2(x-1{)}^2+x^2\\ 3\left(x^2-2\right)-2(x-1{)}^2-x^2<0\\ 3x^2-6-2\left(x^2-2x+1\right)-x^2<0\\ 3x^2-6-2x^2+4x-2-{\mathrm{x̸}}^2<0\\ 4x-8<0\end{array}$

Make it equal to zero,

$$\begin{gathered} 4x-8=0 \\ 4x=8 \\ x=2 \end{gathered}$$

Step 2. Form the intervals

From the obtained values of x, we can form the interval,

So the interval we have is:

$(-\infty ,2)\cup (2,\infty )$

Step 3. Form the table

Since, $f\left(x\right)<0$, so is negative.

check a number in each interval and evaluate the function to see whether it is satisfying the function.

Interval	Number chosen	Resultant
$(-\infty ,2)$	0	$-8<0$
$(2,\infty )$	3	$4<0$

Step 4. Solution set and interval

Since, $-8<0$satisfy $f$. Thus,

role="math" localid="1646209385772" $$\begin{gathered} \left\{x\right|x<2\left\} \\ \right(-\infty ,2) \end{gathered}$$