Solving double inequalities

1. Dec 13, 2009

francis21

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution
1)
LS:
$$3x \leq x+1$$
$$3x-1 \leq x$$
$$-1 \leq -3x$$
$$-1/-2 \geq -2x/-2$$
$$1/2 \geq x$$

RS:
$$x+1 \leq x-1$$
$$0 \leq -2$$

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)
$$x(2x-1) \leq x+7$$
$$2x^2 - x \leq x+7$$
$$2x^2 - 2x - 7 \leq 0$$

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: Dec 13, 2009
2. Dec 13, 2009

tiny-tim

Welcome to PF!

Hi francis21! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

3. Dec 13, 2009

francis21

tiny-tim, I just changed by first post.

I hope this helps.

4. Dec 13, 2009

tiny-tim

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. Dec 13, 2009

francis21

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, $$2x^2-2x-7 \leq 0$$ ($$ax^2+bx +c=0$$), 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. Dec 13, 2009

rootX

I don't think graphing is suggested yet which is easiest way to approach this problem IMO

7. Dec 13, 2009

tiny-tim

Yes, it's a continuous function, so if you find the zeros, you automatically (well, with a little care ) also find the positive and negative ranges.

8. Dec 14, 2009

HallsofIvy

Staff Emeritus
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) $3(0)\le 0+1\le 0-1$
Is that true?

2) $0(2(0)-1)\le 0+ 7$
Is that true?

9. Dec 14, 2009

Mentallic

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 $y\leq 0$? 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