# Thread: Out Sick: Practice Problem Help

1. ## Out Sick: Practice Problem Help

Hello, I was hoping to get some help solving these practice equations from my Algebra class. I have been out sick from my class for a week and a half while we worked on square roots and quadratic equations and I have missed everything and have no idea how to solve them. Any help is appreciated, I'll list the problems below! (Also, I'm not sure how to do superscript on here so if something is squared or cubed I will write that next to it in parenthesis.)
Thank you!

x(3x+2)(x-6)=0

3x(squared)=-4x

2x(squared)+x-3=0

x(squared)-2x+2=0

-x(squared)-6x=2

2. ## Re: Out Sick: Practice Problem Help

Originally Posted by Kristeniscrazy
Hello, I was hoping to get some help solving these practice equations from my Algebra class. I have been out sick from my class for a week and a half while we worked on square roots and quadratic equations and I have missed everything and have no idea how to solve them. Any help is appreciated, I'll list the problems below! (Also, I'm not sure how to do superscript on here so if something is squared or cubed I will write that next to it in parenthesis.)
Thank you!

x(3x+2)(x-6)=0

3x(squared)=-4x

2x(squared)+x-3=0

x(squared)-2x+2=0

-x(squared)-6x=2

Solving Quadratic Equations: Solving by Factoring

3. ## Re: Out Sick: Practice Problem Help

for the first one, use the null factor law, for the rest factor them, then apply the null facotr law.

4. ## Re: Out Sick: Practice Problem Help

Thank you, but I'm still having trouble factoring these. Especially the 5th one.

5. ## Re: Out Sick: Practice Problem Help

$\displaystyle -x^2-6x=2$

$\displaystyle 0=x^2+6x+2$

Now you can complete the square, take the coeffeicent of x, half it then square it, this will be your new constant term.

What do you get?

6. ## Re: Out Sick: Practice Problem Help

Oh sorry, I meant the fifth one... I'm having trouble finding how the factors go with two. Since 2 is prime and the negative indicates that the factors have to have different signs. Idk, maybe I'm going about it wrong.

EDIT: The FOURTH one!! Sorry...

7. ## Re: Out Sick: Practice Problem Help

I believe I did give some advice for the 5th question

Originally Posted by Kristeniscrazy

x(3x+2)(x-6)=0 1st

3x(squared)=-4x 2nd

2x(squared)+x-3=0 3rd

x(squared)-2x+2=0 4th

-x(squared)-6x=2 5th

8. ## Re: Out Sick: Practice Problem Help

Sorry, I edited the above post. I actually meant the fourth one but I kept typing fifth... On my sheet it is number five so I'm getting confused.

9. ## Re: Out Sick: Practice Problem Help

I'm confused as well!

Not to worry, for the 4th problem you'll have to complete the square. Follow the advice in post #5.

10. ## Re: Out Sick: Practice Problem Help

Okay... I think I've solved 4 & 5.

4. x=1(plus or minus)the square root of 3
5. x=-3(plus or minus) the square root of seven

Is that right? If not perhaps I can show my work and figure out what I did wrong.

11. ## Re: Out Sick: Practice Problem Help

I found the smae for Q5, looks like you are on the right track. Q4 might have complex solutions.

12. ## Re: Out Sick: Practice Problem Help

Thank you so much for your help! I'm starting to understand a bit better. Ok, so for the third one would I do the same except divide out the 2 in the beginning?

13. ## Re: Out Sick: Practice Problem Help

I don't think that would help.

Maybe look for a solution in the form $\displaystyle 2x^2+x-3 = (2x+a)(x+b)$