Thread: I have solved most equations but I do need some help please :)

1. I have solved most equations but I do need some help please :)

We are doing a review of algebra 2 and I really need some help. I have a new teacher and I never learned it the way he taught us. He doesn't let us use calculators. So I was wondering if some of you could please help me with some equations! Solve the following polynomial inequalities.

1. (x+2)^2<25

2. 5z^2+26z+3<-2

3. X^4(x-3)<0

4. -2x^2-4x+2x^3 <0

5. 13x+4+5x^2>-2

If anyone could help me with these it would be great. And yes I have tried and been sitting here for an hour stuck on them so don't think I didn't even try. Thank you!

2. Re: I have solved most equations but I do need some help please :)

$(x+2)^2 < 25$

$(x+2)^2 - 25 < 0$

$[(x+2) - 5][(x+2) + 5] < 0$

$(x-3)(x+7) < 0$

critical values (where the left side equals 0) are $x = 3$ and $x = -7$

solution set is $-7 < x < 3$

for further info, go to the link ...

Pauls Online Notes : Algebra - Polynomial Inequalities

3. Re: I have solved most equations but I do need some help please :)

Originally Posted by Allibvarsity
We are doing a review of algebra 2 and I really need some help. I have a new teacher and I never learned it the way he taught us. He doesn't let us use calculators. So I was wondering if some of you could please help me with some equations! Solve the following polynomial inequalities.

1. (x+2)^2<25

2. 5z^2+26z+3<-2

3. X^4(x-3)<0

4. -2x^2-4x+2x^3 <0

5. 13x+4+5x^2>-2

If anyone could help me with these it would be great. And yes I have tried and been sitting here for an hour stuck on them so don't think I didn't even try. Thank you!
Isolate the squares, take the negative and positive square roots, and deal with the inequalities accordingly.

Here are a couple examples to get you started:

(1.) $(x+2)^2<25$
$-5 (assuming x is real)
$-7

You sure you wrote #2 down correctly? It's gonna get awfully messy when you complete the square.

(3.) $x^4(x-3)<0$
$(x-3)<0$ because x^4 has to be nonnegative
$x<3$

4. Re: I have solved most equations but I do need some help please :)

Hello, Allibvarsity!

$2.\;5z^2+26z+3\:<\:-2$

We have: . $5z^2 + 26z + 5 \:<\:0$

When is . $y \:=\:5z^2 + 25z + 5$ .negative?

The graph is an up-opening parabola: $\cup$
It is negative between its x-intercepts.

$5z^2 + 25z + 5 \:=\:0 \quad\Rightarrow\quad (z + 5)(5z+1) \:=\:0 \quad\Rightarrow\quad z \:=\:\text{-}5,\,\text{-}\tfrac{1}{5}$

Therefore: . $\text{-}5\;<\;x\;<\;\text{-}\tfrac{1}{5}$

This technique can be applied to #5.

5. Re: I have solved most equations but I do need some help please :)

For (2), write the inequality as $5x^2+ 26x+ 5< 0$. You can use the quadratic formula to find the roots of the equation, [tex]5x^2+ 26x+ 5= 0, a and b, and then factor as 5(x- a)(x- b)< 0. Of course, such a product will be negative if and only x- a and x- b have opposite signs. And that means x is larger than one of a and b and less than the other.

(4), $2x^3- 2x^2- 4x< 0$ can be factored as $2x(x^2- x- 2)< 0$. And it is easy to factor $x^2- x- 2$. The product of three factors, (x- a)(x- b)(x- c) is negative if all three factors are negative or if two are positive and one negative.

(5), $13x+4+5x^2>-2$ or $5x^2+ 13x+ 6> 0$, can be solved by using the quadratic formula to solve the equation $5x^2+ 13x+ 6= 0$ for the roots a and b. And then the problem can be written as $5(x- a)(x- b)> 0$. That will be true if and only if x-a and x- b have the same sign- in other word if x is larger than both a and b or x is less than both a and b.