Question

@mathmale Thank you for your help! I am try to find the real or imaginary solutions of a polynomial equation. I worked it down to what I think is right but got stuck in trying to find factors of 400 that equal -40.

Answer 1

-20*-20=400
-20 + -20 = -40

Answer 2

Opps! I meant -41, So sorry.

Answer 3

show me your full quadratic please

Answer 4

Let me paraphrase that. We want to find two factors of 400 whose sum is -40.
ehuman has just done that. Ooops. Nicole, I don't believe there are any nice, simple factors of 400 whose sum is -41.

Answer 5

Okay, is another way to find what I am looking for? Would you like me to put up what work I did in trying to solve this?

Answer 6

won't hurt to check that you are correct before moving on.

Answer 7

I'm with ehuman in suggesting that he and I could possibly tell you more if we could see the quadratic equation that you're trying to factor. Yes, there is a straightforward method for determining the two factors you want. I'd be glad to go over that.

Answer 8

Okay, thank you and I will put it up in one moment.

Answer 9

I'll leave it in the capable hands of mathmale so I can play legos with with my son :)
I'll check in soon.

Answer 10

Nice having a son to play LEGO with! Hope you two enjoy your Lego adventures.

Answer 11

Same one as in my avatar. He's 2 weeks old in the pic, 3 years old now.

Answer 12

X^4 – 41x^2 = -400
Plus 400 on both sides
X^4 – 41x^2 + 400 = 0
Use a to equal x^2
A^2 – 41a + 400 = 0
(a + or - _ ) (a + or - _ ) = 0
The last bit ^ right there is where I stop trying to find factor that times to 400 and add/subtract to -41

Thank you! @ehuman

Answer 13

Nicole, are you willing to entertain the possibility that the roots (or zeros) of a^2-41a+400 are not "nice," or, in other words, not integers?

Answer 14

So I would have to use "i" ?

Answer 15

Not necessarily. I encourage you to use the quadratic formula to find two possible values for your a (remembering that a=x^2). Would you consider doing this?

Answer 16

Yes, I'll try right now. Just one moment - I am trying to find the page I was using before(in my book) to see if I possibly had the wrong one.

Answer 17

Okay, I didn't have the wrong one but you may just know a better way(: Thank you again and would you like me to do the quadratic formal from the point a^2 - 41a + 400 = 0?

Answer 18

I've done that in the meantime. The roots x happen to be real and positive.

Answer 19

I'm going to give you a choice: should we talk about how we'd find factors of 400 that sum up to -40, OR continue with the present problem, OR tackle a new one? Your call.

Answer 20

The present one and the 400 to -41 are the same and I would like to stick with figuring them out

Answer 21

OK. I've used the quadratic equation to find solutions for a^2-41a+400. As before, the solutions I've found are real and positive. This means that you could take either solution and set it equal to x^2 and find values of x that satisfy the equation. However...those x values would be pretty ugly...no integers!

Answer 22

So: want to try using the quadratic formula yourself to find those roots of a^2-41a+400, or want to borrow mine?

your call.

Answer 23

Is this a way of doing it?

Answer 24

I didn't want to keep going with it, if it was wrong but if it is right I'll finish it right now.

Answer 25

Yes, it is. One error there...under the radical, that b^2 is (-41)^2, not -41^2.
If you continue with this you'll get the same answer as I. Smart to finish it so that you'll know how to do it on your own next time.

Answer 26

That "b^2" really means "(b)^2"...whether b itself is + or -, you must square the whole thing, sign and all.

Answer 27

Okay - I'll finish it.

Answer 28

they actually end up factoring, I missed it at 1st.

Answer 29

Once you get to square root 81 do you take its root 9 and leave it has 9/2?

Answer 30

*as

Answer 31

ehuman: have you really found two factors of 400 that sum up to -41? Nicole, I'll be right back.

Answer 32

Okay.

Answer 33

I'm afraid I owe you both (ehuman and Nicole) an apology. I tried squaring -41 in my head and got 1641, which is just plain wrong. ehuman helped me to focus on this, with the result that I can now say that under the radical sign we should have 1681-1600.

Answer 34

Therefore, Nicole, our roots are

Answer 35

-25*-16 = 400
-25+ -16 = -41

Answer 36

\[x=\frac{ -41+9 }{2 }\]

Answer 37

and the other root that stems from a (-) sign in front of the 9.
Nicole? What are these 2 roots?

Answer 38

Wow, you tried doing that in your head? I could never. Would it be 41 and not 41- because "b" is negative to start with?

Answer 39

Nicole, you're absolutely right. My bad. My very bad.

Answer 40

+/- 5
and +/-4

Answer 41

Ha, It's okay. Is one of the x's 25?

Answer 42

remember your substitution

Answer 43

YES! in answer to y our question, Nicole.
YES! in answer to your reminder, ehuman.

Answer 44

Okay, so how did @ehuman get the 16?

Answer 45

So, Nicole, as you've probably already figured out for yourself, and which ehuman has done, our a = 25 = x^2. What are the resulting roots?

Answer 46

I'll let ehuman explain that, since he obviously knows his stuff. But I'm sticking around.