Is 0 a number in the solution set?

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Homework Help Overview

The discussion revolves around determining whether 0 is included in the solution set for two inequalities: 3x ≤ x + 1 ≤ x - 1 and x(2x - 1) ≤ x + 7. Participants are exploring the implications of these inequalities and the conditions under which 0 might satisfy them.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the left side of the first inequality and conclude that it suggests 0 is not in the solution set. They also explore the right side, noting it yields no solutions. For the second inequality, attempts to factor and analyze the quadratic expression are mentioned, alongside a suggestion to substitute 0 directly into the inequalities to check validity.

Discussion Status

There is an ongoing exploration of different methods to assess whether 0 is part of the solution set. Some participants suggest substituting 0 directly into the inequalities, while others focus on solving the inequalities to understand the ranges of solutions. Multiple interpretations of the problem are being discussed without a clear consensus.

Contextual Notes

Participants note that the homework instructions specifically ask whether 0 is in the solution set, rather than requiring a full solution to the inequalities. This has led to discussions about the nature of the inequalities and the approach to determining the solution set.

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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.
 
Last edited:
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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:
 
tiny-tim, I just changed by first post.

I hope this helps.
 
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?
 
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)?
 
I don't think graphing is suggested yet which is easiest way to approach this problem IMO
 
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.
 
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?
 
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:
 

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