Question

How many real solutions does the equation x^2 + 4x - n^2 = 0 have?

A.None
B. One
C. Two
D. Cannot be determined

Also, we are learning how to choose "Cannot be determined" properly. So make sure 100% that you can't solve it, before answering cannot be determined! Thanks

Answer 1

C. Two

Answer 2

Explanation?

Answer 3

This is a quadratic equation. Apparently, all the equations with the degree as '2' have two solutions.

Answer 4

@CoCoTsoi will love to give tat!

Answer 5

discriminant = b^2 - 4ac..here its 16+4n^2
now here assuming n is a real no,,can you tell me whether discriminant is greater than 0,less than it or maybe equal to it.?

Answer 6

To check if a quadratic has real factors (solutions) or not, check the Discriminant "D"
\[D = b^2 - 4ac\]

where a,b and ccan be found by comparing your quadratic with the standard form:
\[ax^2 + bx +c = 0\]

now, if:

D=0. --->One real root or factor (that means its a complete square)
D>0 --->Two real factors
D<0 ---> NO real factors

So, what do you think now for the above problem?

Answer 7

For some reason I got -15... which according to @apoorvk is no real factors, something must be wrong hmmm

Answer 8

Answer is C. Why, let me show you
First tell me this - what would a,b,c in this case be?

Answer 9

i worked it again. For some reason I thought the original equation wasn't in standard form (is it?).. I got 20 this time.

Answer 10

20 (D) > 0, so would it be c?

Answer 11

a would be 1, b would be 4, and c would be -1? @apoorvk

Answer 12

how can you get 20 or some other constant when you dont know n..

Answer 13

c would be "-n^2"

Answer 14

think..

Answer 15

c= -n^2 .. just treat it like any other constant real no..

Answer 16

@shubhamsrg well I assumed if there is a - sign there is a 1 in front of n? is that stupid?

Answer 17

Are there any conditions for n?

Answer 18

no that isn't stupid.

but the whole constant term is "-n^2" right? that 'n^2' is definitely not a variable

Answer 19

well n is not a variable and as i said before..treat is like any other constant no.

Answer 20

ok...

Answer 21

I guess too many cooks are spoiling the broth here, so should I leave it to you guys at the moment?

Answer 22

No no no.. I still need help from everyone. I'm still a little confused

Answer 23

okay, np

Answer 24

x^2+4x-n^2=0

4^2-4(1)(-n^2)
=16+4n^2

if there is two roots,
16+4n^2>0
4n^2>-16
n^2 >-4
n>2i or n<-2i

if there is one root,
16+4n^2=0
n=2i or -2i

if there is no root,
16+4n^2<0
-2i<n<2i

that means if n can be all values, we can get different numbers of roots.
I would choose cannot determine

Answer 25

@CoCoTsoi what is say is true..but at this standard i dont think they want us to consider n as a complex no..

Answer 26

i mean what you say is true *

Answer 27

That's why I asked if there is condition for n :P

Answer 28

keep in mind this is an algebra 1 class

Answer 29

I would still choose cannot be determined as we do not know what n is :D

Answer 30

@apoorvk You think it's c? Can you show me the steps? I'm kinda lost? Who else things it's c, two solutions?

Answer 31

Just tell me something, Kiera - just soppose that 'n' is any constant. so n62 would be any constant - right?

In place of 'c' in ax^2 +bx +c in your equation, we have '-n^2' right?

Answer 32

*suppose and & **n^2

Answer 33

yes

Answer 34

So what would the discriminant be?

Answer 35

ok what would the c be in the discriminant?

Answer 36

would it be like -1 or just -n?

Answer 37

@Rohangrr copied the answer from here: http://answers.yahoo.com/question/index?qid=20100630112659AA1OiUz

Answer 38

D = b^2 - 4ac = 4^2 - 4(1)(-(n^2)) = 16 + 4n^2 ---> am i right?

Answer 39

@apoorvk Yes

Answer 40

@FoolForMath nothing new with that - Y! Answers is @Rohangrr 's favorite.


@Callisto i was asking the kid - lol.

Answer 41

@KieraJan Have you learnt complex number yet?

Answer 42

@ @KieraJan
suppose the equation would have been "x^2 + 4x - 6" - what would have the 'c' be been?

Answer 43

no @Callisto we are making it too complicated - this is really a stupid problem - there is no complex no. involved.

Answer 44

@Callisto no

Answer 45

@KieraJan - I asked - suppose the equation would have been "x^2 + 4x - 6" - what would have the 'c' been in this case??

Answer 46

D = b^2 - 4ac = 4^2 - 4(1)(-(n^2)) = 16 + 4n^2 - I followed that @apoorvk

Answer 47

-6 right?

Answer 48

very well - that means we are done!! well almost..


so now tell me - can n^2 be negative ever?

Answer 49

I'm not sure...

Answer 50

i guess it was in this problem?

Answer 51

No, don't guess - how did you arrive at '-6'?

Answer 52

because there was a negative in front of 6, the c.. so -6

Answer 53

so if there is a negative in front of n^2, I'm thinking it could probably be a negative

Answer 54

Gorget '-6'. A whole square can NEVER be negative. because a positive no. multiplied by itself is always positive, as well as a negative no. multiplied by itself is postive - because two '-' signs make a plus sign - right?

for g.--> (-4)^2 = -4 x -4 = +16

Answer 55

yes you are right,

Answer 56

yes because to negatives make a positive

Answer 57

yeah. but the whole equation was---> 4^2 - (- 4xn^2) --> right?? adjacebt minus sign makes a plus right? so it transforms to --> 16 '+' 4n^2 - right?

Answer 58

yes

Answer 59

that means that, whatever 'n' is - negative OR positive, n^2 will always be positive - right?

Answer 60

right!

Answer 61

and if n^2 is positive, then 4n^2 would also be positive - positive multiplied by a positive is positive I heard. - okay?

Answer 62

yes

Answer 63

so we have 16 + 4n^2

Answer 64

and now, 16 added to 4n^2, which is "16 + 4n^2" is also positive - since positive added to positive is positive as well?

Answer 65

yes, ,we have 16+4n^2, which is the DISCRIMINANT, which is positive - ALWAYS!

Answer 66

and you said D>0 is two solutions.. so how does that tie in to it?

Answer 67

the minimum - D can be in this case is when n=0, that is D is atleast 16. so always greater 0 anyhow - so two solutions.

Answer 68

so to show my work could I do: 16 + 4n^2 = 20n^2 which 20n^2 > 0 , therefor two solutions (answer choice C) @ apoorvk

Answer 69

NO!!

16 + 4n62 is that itself - how can you add two terms '16" and "4n^2"?
does "2x + 4' equal "6x"? NO - similarly here as well.

Answer 70

so I could say 16 + 4n^2 > 0 , therefor two solutions?

Answer 71

**(16 + 4n^2)

You just let it remain that way.
As I said each term is always positive -
"16" is positive.
"4n^2" is positive - always - as we derived it above^.

so positive plus positive is also positive.

hence the whole "16 + 4n62" is positive too.

Answer 72

yeah.

Answer 73

so either way since neither is negative, even when added together, it won't go below 0, therefor it will always be two solutions, greater than 0?

Answer 74

Exactly!!!! Now you're getting it!

Answer 75

ah ha! (:

Answer 76

so if you were doing my homework, you would agree that it was C, two solutions? and not Cannot be determined?

Answer 77

hundred percent.

Answer 78

good!

Answer 79

So I now solved the problem completely, I think?

Answer 80

yup - just follow up with the correct statements - you have done that I hope?

Answer 81

Thanks to everyone for your help! (: yes I have

Answer 82

very good. no worries :)