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 x^2- 3x+ 1has roots \frac{3\pm\sqrt{5}}{2}. This time the inequality is satisfied for x &lt;b&gt;outside&lt;/b&gt; those points. \frac{3+ \sqrt{5}}{2}is approximately 2.6, &amp;amp;lt;b&amp;amp;gt;less&amp;amp;lt;/b&amp;amp;gt; 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}.

 

[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 x^2- 3x+ 1has roots \frac{3\pm\sqrt{5}}{2}. This time the inequality is satisfied for x &lt;b&gt;outside&lt;/b&gt; those points. \frac{3+ \sqrt{5}}{2}is approximately 2.6, &amp;amp;lt;b&amp;amp;gt;less&amp;amp;lt;/b&amp;amp;gt; 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}.

&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; 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?

 

[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 x^2- 3x+ 1has roots \frac{3\pm\sqrt{5}}{2}. This time the inequality is satisfied for x &lt;b&gt;outside&lt;/b&gt; those points. \frac{3+ \sqrt{5}}{2}is approximately 2.6, &amp;amp;lt;b&amp;amp;gt;less&amp;amp;lt;/b&amp;amp;gt; 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}.

&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; I made a typo. There should be no x in front of the absolute value.

 

[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?