Answer:
(21) -5 < x < 0
(22) [tex]\displaystyle\bf-\frac{8}{3} < x < 2[/tex]
Step-by-step explanation:
We can find the solutions for x² < -5x and |3x + 1| < 7 by using these ways:
(21)
Steps to solve a quadratic inequality:
<1> Move all the terms to 1 side, so the other side is 0:
[tex]\begin{aligned}x^2 & < -5x\\x^2+5x& < 0\end{aligned}[/tex]
<2> Factorize the quadratic inequality:
[tex]\begin{aligned}x^2+5x& < 0\\x(x+5)& < 0\end{aligned}[/tex]
<3> Determine the root(s):
[tex]\texttt{for }x(x+5)=0\texttt{, }x=0\texttt{ or }-5[/tex]
<4> Find the range that meet the inequality condition by using a number line or a graph (see the picture):
Since the inequality is less than 0, then the range that meets the condition is -5 < x < 0.
(22)
Steps to solve an inequality with 1 absolute value:
<1> move the absolute value to 1 side and the rest to the other side (which is already done in this question)
<2> check the following conditions:
- If |f(x)| < c, then -c < f(x) < c
- If |f(x)| > c, then f(x) < -c or f(x) > c
In this question:
[tex]|3x+1| < 7[/tex]
then:
[tex]-7 < 3x+1 < 7[/tex]
<3> solve the inequality:
first we solve -7 < 3x + 1, next we solve 3x + 1 < 7, then the solution is the range that meets both inequalities.
[tex]-7 < 3x+1[/tex]
[tex]3x > -7-1[/tex]
[tex]x > -8\div3[/tex]
[tex]\displaystyle\bf x > -\frac{8}{3}[/tex]
[tex]3x+1 < 7[/tex]
[tex]3x < 7-1[/tex]
[tex]x < 6\div3[/tex]
[tex]\bf x < 2[/tex]
The range that meets [tex]\bf x > -\frac{8}{3}[/tex] and [tex]\bf x < 2[/tex] is:
[tex]\displaystyle\bf-\frac{8}{3} < x < 2[/tex]
