Is 0 a number in the solution set?

In summary: For 1) you could try graphing it to see if it looks like the equation suggests it should. But again, since this is a linear inequality, graphing might not be 100% necessary. Just try solving for x.For number 2), 0(2(0)-1)\le 0+ 7, so y\leq 0 for x=-1, 0, 1. In other words, the parabola is less than or equal to 0 at the points (-1,0), (0,1).
  • #1
francis21
8
0

Homework Statement



Instructions: For the following inequalities, determine if 0 is a number in the solution set.
1) 3x [tex]\leq[/tex] x+1 [tex]\leq[/tex] x-1
2) x(2x-1) [tex]\leq[/tex] x+7

Homework Equations





The Attempt at a Solution


1)
LS:
[tex]3x \leq x+1 [/tex]
[tex]3x-1 \leq x [/tex]
[tex]-1 \leq -3x [/tex]
[tex]-1/-2 \geq -2x/-2[/tex]
[tex]1/2 \geq x [/tex]

RS:
[tex]x+1 \leq x-1[/tex]
[tex]0 \leq -2[/tex]

I said for the right side that there are no solutions, since the inequality is not true.

So I said that 0 is not included in the solution set in this inequality, based from the inequality from the left side.

2)
[tex] x(2x-1) \leq x+7 [/tex]
[tex] 2x^2 - x \leq x+7 [/tex]
[tex] 2x^2 - 2x - 7 \leq 0 [/tex]

After that I tried factoring the equation. It would work, just not with even numbers though...have some decimals on it.

This is what I tried so far for both questions.

Note: I looked at the wrong part of the question, that's why I changed the question, but its still the same equations.
 
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  • #2
Welcome to PF!

Hi francis21! Welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
tiny-tim, I just changed by first post.

I hope this helps.
 
  • #4
francis21 said:

Homework Statement



Instructions: For the following inequalities, determine if 0 is a number in the solution set.
1) 3x [tex]\leq[/tex] x+1 [tex]\leq[/tex] x-1
2) x(2x-1) [tex]\leq[/tex] x+7

Homework Equations





The Attempt at a Solution


1)
LS:
[tex]3x \leq x+1 [/tex]
[tex]3x-1 \leq x [/tex]
[tex]-1 \leq -3x [/tex]
[tex]-1/-2 \geq -2x/-2[/tex]
[tex]1/2 \geq x [/tex]

RS:
[tex]x+1 \leq x-1[/tex]
[tex]0 \leq -2[/tex]

I said for the right side that there are no solutions, since the inequality is not true.

So I said that 0 is not included in the solution set in this inequality, based from the inequality from the left side.

2)
[tex] x(2x-1) \leq x+7 [/tex]
[tex] 2x^2 - x \leq x+7 [/tex]
[tex] 2x^2 - 2x - 7 \leq 0 [/tex]

After that I tried factoring the equation. It would work, just not with even numbers though...have some decimals on it.

This is what I tried so far for both questions.

Note: I looked at the wrong part of the question, that's why I changed the question, but its still the same equations.

1) looks ok, and for 2), use the standard formula for a quadratic equation …

except why not just substitute x = 0 into the inequalities, and see if they're true?
 
  • #5
tiny-tim said:
1) looks ok, and for 2), use the standard formula for a quadratic equation …

except why not just substitute x = 0 into the inequalities, and see if they're true?

That would actually make sense. But I guess, for this particular homework, they want to train us on how to solve linear inequalities. That's why I was solving for the inequalities.

For number 1, I tried placing zero on the inequality, and does not actually include zero as its solution in the solution set.

For number 2, isn't written in standard form, [tex] 2x^2-2x-7 \leq 0 [/tex] ([tex] ax^2+bx +c=0[/tex]), but just expressed as an inequality?

But let's say if I factor the quadratic equation, the x values would be the interval right (x= some value or x= some other value)?
 
  • #6
I don't think graphing is suggested yet which is easiest way to approach this problem IMO
 
  • #7
francis21 said:
But let's say if I factor the quadratic equation, the x values would be the interval right (x= some value or x= some other value)?

Yes, it's a continuous function, so if you find the zeros, you automatically (well, with a little care :wink:) also find the positive and negative ranges.
 
  • #8
francis21 said:

Homework Statement



Instructions: For the following inequalities, determine if 0 is a number in the solution set.
1) 3x [tex]\leq[/tex] x+1 [tex]\leq[/tex] x-1
2) x(2x-1) [tex]\leq[/tex] x+7
Notice that this does NOT ask you to solve the inequality. It only asks whether "0 is a number in the solution set"- that is, whether or not 0 satisfies the inequality. Just replace x by 0 in each and see if they are true.

1) [itex]3(0)\le 0+1\le 0-1[/itex]
Is that true?

2) [itex]0(2(0)-1)\le 0+ 7[/itex]
Is that true?
 
  • #9
HallsofIvy said:
Notice that this does NOT ask you to solve the inequality. It only asks whether "0 is a number in the solution set"- that is, whether or not 0 satisfies the inequality. Just replace x by 0 in each and see if they are true.

1) [itex]3(0)\le 0+1\le 0-1[/itex]
Is that true?

2) [itex]0(2(0)-1)\le 0+ 7[/itex]
Is that true?
francis21 said:
That would actually make sense. But I guess, for this particular homework, they want to train us on how to solve linear inequalities. That's why I was solving for the inequalities.

For 2) try drawing the parabola. It will make more sense this way.
Now, factorize (or use the quadratic formula, whichever suits your fancy) to find the x-intercepts, and now try to see where the parabola is less than or equal to 0. In other words, for what range of x-values is [itex]y\leq 0[/itex]? If the x-value of 0 is in this range, then 0 is in the solution set of the inequality.

p.s. quadratic inequalities aren't linear :cool:
 

Related to Is 0 a number in the solution set?

1. What is a double inequality?

A double inequality is an inequality that contains two inequality symbols, such as x < 3 and x > 1. This means that the value of x must be greater than 1 and less than 3 to satisfy the inequality.

2. How do you solve a double inequality?

To solve a double inequality, you must first isolate the variable on one side of the inequality. Then, you can solve each inequality separately to find the range of values that satisfy both inequalities.

3. Can you use the same rules for solving single inequalities?

Yes, the same rules for solving single inequalities can be used for solving double inequalities. However, you must pay attention to the direction of the inequality symbols and make sure to solve both inequalities simultaneously.

4. What is the difference between solving a double inequality and graphing a double inequality?

Solving a double inequality involves finding the range of values that satisfy the inequality. Graphing a double inequality involves representing this range of values on a number line or coordinate plane.

5. Are there any special cases to consider when solving double inequalities?

Yes, when multiplying or dividing both sides of a double inequality by a negative number, the direction of the inequality symbol must be reversed. This is because multiplying or dividing by a negative number flips the inequality. For example, 2x > -4 becomes x < -2 when both sides are divided by -2.

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