Question

solve this polynomial equation:
2x^4 - 10x^2 = 72

Answer 1

let \(\large\color{black}{ x^2 =a }\)

Answer 2

re-write your equation in terms of a, and solve it.

Answer 3

you can also factor out 2 to work with smaller numbers

Answer 4

that's not how my math teacher taught us how to do it so i dont know what you mean. for this equation, we'd have to set it to 0 so we'd have to subtract 72 from both sides

Answer 5

subtracting 72 from both sides, \(\large\color{black}{ 2x^4 - 10x^2-72\color{red}{-72} = 0\color{red}{-72} }\)
this will get you, \(\large\color{black}{ 2x^4 - 10x^2-72 = 0 }\)

then say, let \(\large\color{magenta}{ x^2=a }\)
lets do our substitution,

\(\large\color{black}{ 2x^4 - 10x^2-72 = 0 }\)
\(\large\color{black}{ 2(x^2)^2 - 10x^2-72 = 0 }\)
\(\large\color{black}{ 2(\color{magenta}{x^2})^2 - 10\color{magenta}{x^2}-72 = 0 }\)
\(\large\color{black}{ 2(\color{magenta}{a})^2 - 10\color{magenta}{a}-72 = 0 }\)

Answer 6

So, you get, \(\large\color{black}{ 2a^2 - 10a-72 = 0 }\)

Answer 7

Can you solve this equation?

Answer 8

honestly, I don't know what to do. I see it and my mind is blank. my dad is currently trying to help me and he's usually really good at math. he doesnt know what the heck to do

Answer 9

You, right now, have to worry just about solving this, \(\Large\color{black}{2a^2-10a-72=0 }\)

Answer 10

\(\Large\color{black}{2(a^2-5a-36)=0 }\)
\(\Large\color{black}{a^2-5a-36=0 }\)
So far so good?

then, I'll give you a hint, \(\Large\color{black}{36=\color{red}{9}\times \color{darkgoldenrod}{4} }\) and \(\Large\color{black}{\color{red}{9}-\color{darkgoldenrod}{4}=5 }\)
use this to think how to factor your equation.

Answer 11

this is what I did for the equation, which was on our test. we're allowed to do test corrections and I got 4 points off, and I don't know why. we also had to find the x values and I got -4 and 9 for it. what did I do wrong?

Answer 12

ohh, these are not the `x` values, you did,
\(\large\color{black}{ a^2-5a-36=0 }\)
\(\large\color{black}{(a-9)(a+4)=0 }\)
\(\large\color{black}{a=-4,~~9 }\) right?

Answer 13

but, we initially said, that \(\large\color{magenta}{a=x^2 }\) , so there was an additional step.

\(\large\color{black}{x^2=9 }\) or \(\large\color{black}{x^2=-4 }\)

Answer 14

yes that's what I did. so would the x's be 3 and -2?

Answer 15

No.

Answer 16

Have you learned about imaginary number, \(\large\color{black}{i }\) ?

Answer 17

yes i have

Answer 18

Okay, I can see you haven't...
\(\large\color{black}{x^2=-4 }\) will have 2 imaginary solutions, \(\large\color{black}{x=\pm2i }\), but no real solutions.

Answer 19

ohh, you have... sorry.

Answer 20

So you can see why I am getting these soltuions? right?

Answer 21

yeah I see. and 9 would then become x= \[\pm 3i\]

Answer 22

NO.

Answer 23

calm down, please. I am trying to understand. not everybody is a math wiz

Answer 24

you have, \(\large\color{black}{x^2=9 }\) , and not, \(\large\color{black}{x^2=\color{red}{-}9 }\)

Answer 25

I just said, no. I wasn't harsh, or anything, was I?

Answer 26

it kinda seemed like it. but don't worry about it. what would we do with 9 then?

Answer 27

oh I forgot to say this before. I got -4 and 9 because those are what you'd need to get 4 and -9 to equal 0. like for (x-2)(x+2)(x+3), x would equal \[\pm2\] and -3.

Answer 28

@mcaMohawk @SolomonZelman was not being mean or harsh etc. . He just needs you to stay focused so you can really get it. you can be a math whiz!
In your explanation you mean you got -4 and 9 because that's what you need to get -5