How to Solve the Absolute Value Inequality x|x-3|<-1 for 3x+1?

[mathdad]

See question for question.

View attachment 8539

 

[HOI]

The first thing I would do is remove the "absolute value". If x> 3 then x- 3>0 so |x- 3|= x- 3. The inequality becomes $$x(x- 3)= x^2- 3x\le 1$$ We can write that as $$x^2- 3x- 1\le 0$$.

The equation, $$x^2- 3x- 1= 0$$ has roots $$x= \frac{3\pm\sqrt{13}}{2}$$. This is a parabola that opens upward so the inequality is satisfied for x between those two points. Taking the positive sign, $$\frac{3+ \sqrt{13}}{2}$$ is about 4.67, just slightly larger than 3. Since we require here that x> 3, we require that $$3\le x\le \frac{3\pm\sqrt{13}}{2}$$.

If x< 3, |x- 3| is negative so $$x|x- 3|= -x(x- 3)= 3x- x^2$$. The inequality becomes $$3x- x^2\le 1$$ or $$x^2- 3x+ 1\ge 0$$. The equation [tex]x^2- 3x+ 1[tex]has roots $$\frac{3\pm\sqrt{5}}{2}$$. This time the inequality is satisfied for x <b>outside</b> those points. [tex]\frac{3+ \sqrt{5}}{2}[tex]is approximately 2.6, <b>less</b> than 3. Of course $$\frac{3- \sqrt{5}}{2}$$ is less than 3. So this inequality is satisfied by $$x\le \frac{3- \sqrt{5}}{2}$$ and $$\frac{3+\sqrt{5}}{2}\le x\le \frac{3\pm\sqrt{13}}{2}$$.[/tex][/tex][/tex][/tex]

 

[Wilmer]

I seek the first 2 steps.

Step 1: get paper
Step 2: get pencil with eraser

 

[mathdad]

The first thing I would do is remove the "absolute value". If x> 3 then x- 3>0 so |x- 3|= x- 3. The inequality becomes $$x(x- 3)= x^2- 3x\le 1$$ We can write that as $$x^2- 3x- 1\le 0$$.

The equation, $$x^2- 3x- 1= 0$$ has roots $$x= \frac{3\pm\sqrt{13}}{2}$$. This is a parabola that opens upward so the inequality is satisfied for x between those two points. Taking the positive sign, $$\frac{3+ \sqrt{13}}{2}$$ is about 4.67, just slightly larger than 3. Since we require here that x> 3, we require that $$3\le x\le \frac{3\pm\sqrt{13}}{2}$$.

If x< 3, |x- 3| is negative so $$x|x- 3|= -x(x- 3)= 3x- x^2$$. The inequality becomes $$3x- x^2\le 1$$ or $$x^2- 3x+ 1\ge 0$$. The equation [tex]x^2- 3x+ 1[tex]has roots $$\frac{3\pm\sqrt{5}}{2}$$. This time the inequality is satisfied for x <b>outside</b> those points. [tex]\frac{3+ \sqrt{5}}{2}[tex]is approximately 2.6, <b>less</b> than 3. Of course $$\frac{3- \sqrt{5}}{2}$$ is less than 3. So this inequality is satisfied by $$x\le \frac{3- \sqrt{5}}{2}$$ and $$\frac{3+\sqrt{5}}{2}\le x\le \frac{3\pm\sqrt{13}}{2}$$.[/tex][/tex][/tex][/tex]

[tex][tex][tex]$$<br /> I cannot read your reply on my phone. The words block most of the LaTex. In any case, what is the value of a and b? Are you saying that the value of a and b is the fraction involving the sqrt{13} on the numerator?$$[/tex][/tex][/tex]

 

[Joppy]

I cannot read your reply on my phone. The words block most of the LaTex.

Is this better?

The first thing I would do is remove the "absolute value". If x> 3 then x- 3>0 so |x- 3|= x- 3. The inequality becomes $$x(x- 3)= x^2- 3x\le 1$$ We can write that as $$x^2- 3x- 1\le 0$$.

The equation, $$x^2- 3x- 1= 0$$ has roots $$x= \frac{3\pm\sqrt{13}}{2}$$. This is a parabola that opens upward so the inequality is satisfied for x between those two points. Taking the positive sign, $$\frac{3+ \sqrt{13}}{2}$$ is about 4.67, just slightly larger than 3. Since we require here that x> 3, we require that $$3\le x\le \frac{3\pm\sqrt{13}}{2}$$.

If x< 3, |x- 3| is negative so $$x|x- 3|= -x(x- 3)= 3x- x^2$$. The inequality becomes $$3x- x^2\le 1$$ or $$x^2- 3x+ 1\ge 0$$. The equation $$x^2- 3x+ 1$$ has roots $$\frac{3\pm\sqrt{5}}{2}$$. This time the inequality is satisfied for x outside those points. $$\frac{3+ \sqrt{5}}{2}$$ is approximately 2.6, less than 3. Of course $$\frac{3- \sqrt{5}}{2}$$ is less than 3. So this inequality is satisfied by $$x\le \frac{3- \sqrt{5}}{2}$$ and $$\frac{3+\sqrt{5}}{2}\le x\le \frac{3\pm\sqrt{13}}{2}$$.

In any case, what is the value of a and b? Are you saying that the value of a and b is the fraction involving the sqrt{13} on the numerator?

From the last inequality involving $x$, what do you have to do to $x$ to obtain $3x + 1$ ?

 

[mathdad]

The first thing I would do is remove the "absolute value". If x> 3 then x- 3>0 so |x- 3|= x- 3. The inequality becomes $$x(x- 3)= x^2- 3x\le 1$$ We can write that as $$x^2- 3x- 1\le 0$$.

The equation, $$x^2- 3x- 1= 0$$ has roots $$x= \frac{3\pm\sqrt{13}}{2}$$. This is a parabola that opens upward so the inequality is satisfied for x between those two points. Taking the positive sign, $$\frac{3+ \sqrt{13}}{2}$$ is about 4.67, just slightly larger than 3. Since we require here that x> 3, we require that $$3\le x\le \frac{3\pm\sqrt{13}}{2}$$.

If x< 3, |x- 3| is negative so $$x|x- 3|= -x(x- 3)= 3x- x^2$$. The inequality becomes $$3x- x^2\le 1$$ or $$x^2- 3x+ 1\ge 0$$. The equation [tex]x^2- 3x+ 1[tex]has roots $$\frac{3\pm\sqrt{5}}{2}$$. This time the inequality is satisfied for x <b>outside</b> those points. [tex]\frac{3+ \sqrt{5}}{2}[tex]is approximately 2.6, <b>less</b> than 3. Of course $$\frac{3- \sqrt{5}}{2}$$ is less than 3. So this inequality is satisfied by $$x\le \frac{3- \sqrt{5}}{2}$$ and $$\frac{3+\sqrt{5}}{2}\le x\le \frac{3\pm\sqrt{13}}{2}$$.[/tex][/tex][/tex][/tex]

[tex][tex][tex]$$<br /> I made a typo. There should be no x in front of the absolute value.$$[/tex][/tex][/tex]

 

[mathdad]

Is this better?From the last inequality involving $x$, what do you have to do to $x$ to obtain $3x + 1$ ?

A friend responded to my question this way:

|x - 3| ≤ 1

-1 ≤ x - 3 ≤ 1

3 * (-1) ≤ 3 * (x - 3) ≤ 3 * 1

-3 ≤ 3x - 9 ≤ 3

-3 + 10 ≤ 3x - 9 + 10 ≤ 3 + 10

7 ≤ 3x + 1 ≤ 13

a = 7, b = 3

Is this correct?

Where did 10 come from in his reply?

 

[Joppy]

|x - 3| ≤ 1

-1 ≤ x - 3 ≤ 1

3 * (-1) ≤ 3 * (x - 3) ≤ 3 * 1

-3 ≤ 3x - 9 ≤ 3

-3 + 10 ≤ 3x - 9 + 10 ≤ 3 + 10

7 ≤ 3x + 1 ≤ 13

a = 7, b = 3

Is this correct?

Yes, though it might have been simpler to get $x$ by itself from the first line.

 

[mathdad]

I will try more similar questions at home. If I get the wrong answer or get stuck, I will post all three questions with my work shown.

 

[I like Serena]

I will try more similar questions at home. If I get the wrong answer or get stuck, I will post all three questions with my work shown.

Did you read my previous PM about posting useless comments?

 

[mathdad]

Did you read my previous PM about posting useless comments?

No comment is useless.

 

[Joppy]

No comment is useless.

The statements:

I will try more similar questions at home. If I get the wrong answer or get stuck, I will post all three questions with my work shown.

don't have any value or 'use' to anyone here reading them. It's good that you are trying more questions, but it's a bit like if I told you that this afternoon I'm going to go shopping and buy some milk and bread.

 

[mathdad]

The statements:
don't have any value or 'use' to anyone here reading them. It's good that you are trying more questions, but it's a bit like if I told you that this afternoon I'm going to go shopping and buy some milk and bread.

I totally understand what you're saying. I just thought that revealing some of my daily activities would make for extra conversation. However, I will only stick to math and math only.

 

[mathdad]

A friend responded to my question this way:

|x - 3| ≤ 1

-1 ≤ x - 3 ≤ 1

3 * (-1) ≤ 3 * (x - 3) ≤ 3 * 1

-3 ≤ 3x - 9 ≤ 3

-3 + 10 ≤ 3x - 9 + 10 ≤ 3 + 10

7 ≤ 3x + 1 ≤ 13

a = 7, b = 3

Is this correct?

Where did 10 come from in his reply?