anonymous
  • anonymous
Is this right ? (2x-8)(2x+1)=x^2-16 2(x-4)(2x+1)=(x-4)(x+4) l /x-4 2(2x+1)=x+4 l -x+4 4x+2-x+4=0 3x+6=0 l -6 3x=-6 l /3 x=-2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Yes .-.
anonymous
  • anonymous
seems unlikely but maybe
Hero
  • Hero
Following me around again @satellite73

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I got the same answer , lemme see if i can get a different answer by doing it right ..
anonymous
  • anonymous
no it is not right
anonymous
  • anonymous
@Hero yeah, i need to keep you honest!
anonymous
  • anonymous
Here's how to figure out if you ever doubt yourself , take your number you get for x which in your case you got -2 , & simply substitute it for x .. so it should be (2)(-2)-8(2)(-2)+1 = -2 squared - 16
Hero
  • Hero
I was hoping I would be able to travel to and fro without having to watch my back.
anonymous
  • anonymous
What is it then ?
DebbieG
  • DebbieG
The first problem is here: 2(x-4)(2x+1)=(x-4)(x+4) l /x-4 You can divide by x-4, if you don't know that x-4 is not=0.
Hero
  • Hero
Use quad formula, complete the square or whatever method you feel comfortable with to isolate x.
anonymous
  • anonymous
oops
DebbieG
  • DebbieG
Compounded here: 2(2x+1)=x+4 l -x+4 4x+2-x+4=0 IF the step before were correct, you would subtract (x+4), not subtract x and add 4.
anonymous
  • anonymous
\(4x^2-x^2=?\)
DebbieG
  • DebbieG
*You CAN'T divide by x-4, if you don't know that x-4 is not=0.
anonymous
  • anonymous
@Hero did i make you nervous or something?
Hero
  • Hero
No, you didn't, I just goofed.
Hero
  • Hero
(2x−8)(2x+1)=x^2−16 Multiply the right side: 2x(2x+1)−8(2x+1)=x^2−16 4x^2+2x−16x−8=x^2−16 4x^2−14x−8=x^2−16 Subtract x^2 from both sides: 3x^2−14x−8=−16 Add 16 to both sides: 3x^2−14x+8=0
Hero
  • Hero
Now, you can use quad formula, or complete the square, or whatever method feels comfortable for you.
anonymous
  • anonymous
I need to know what x is.
Hero
  • Hero
Do you know any methods to isolate x such as factoring, quadratic formula or complete the square?
anonymous
  • anonymous
I know methods to facrorise.
Hero
  • Hero
Well, believe it or not 3x^2 -14x + 8 = 0 Is factorable
anonymous
  • anonymous
a^2-2ab+b^2=(a-b)^2 ?
Hero
  • Hero
That only works if you have a perfect square.
anonymous
  • anonymous
Yes.
Hero
  • Hero
3x^2 - 14x + 8 = 0 that's not a perfect square, but it is factorable. Try finding two numbers that multiply to get 24 yet add to get -14 m x n = 24 m + n = -14
anonymous
  • anonymous
12 and 2.
Hero
  • Hero
Actually -12 and -2
anonymous
  • anonymous
Yes.
Hero
  • Hero
We can replace -14 with -(12 + 2) in the equation: 3x^2 - (12 + 2)x + 8 = 0 3x^2 -12x - 2x + 8 = 0
Hero
  • Hero
Now we can factor by grouping.
Hero
  • Hero
If you factor the first two terms you get 3x(x - 4) If you factor the last two terms you gete -2(x - 4) So you have 3x(x - 4) - 2(x - 4) = 0 x - 4 is also common to both so you factor that out to get (x - 4)(3x - 2) = 0 From here, you use zero product property: x - 4 = 0 3x - 2 = 0 So there are two values of x

Looking for something else?

Not the answer you are looking for? Search for more explanations.