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# help

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Find the number of quadratic equations of the form $$x^2 + ax + b = 0,$$ such that whenever $$c$$ is a root of the equation,$$c^2 - 2$$ is also a root of the equation.

Jan 19, 2019

#1
+5226
+3

$$x^2+ax+b = (x-c)(x-(c^2-2) = \\ (x-c)(x-c^2+2) = \\ x^2 +(2-c-c^2)x + (c^3-2c) \\$$

$$\text{As we don't have any limitations on }c\\ \text{There are infinitely many equations where both }c \text{ and }c^2-2\\ \text{are both roots, given by }x^2+ax+b \\ a = 2-c-c^2\\b = c^3-2c$$

.
Jan 19, 2019
#2
+533
0

this means that a=-c^2-c+2 (vieta's) and b=c^3-2c (also vieta's)

now you can see that all you need to do is plug in any value for c to get any two values for a and b. this means that there are infinitely many solutions to the equation.

Jan 19, 2019
#3
-2

You are both incorrect.

Jan 19, 2019
#14
+5226
+2

if you're going to waste our time with some trick question please state it ahead of time.

your life may be so invaluable that you have time to waste on b******t, mine isn't.

Rom  Jan 19, 2019
#15
+102461
+2

If you have reason to think the answers are wrong then you can say so politely and you should give the reason that you think this.

I have looked at Rom's answer and it looks good to me.

Asdf has asked someone to take a look.

Hi asdf,

I do not know what  "vieta's" means and I do not understand your response.

I suggest you look at what Rom did.

asdf, when guests gives the  response of "this is wrong"  you are best to counter with the response "Please explain why you think so".

Encourage the guest to elaborate.

That is if you really care about it.

Melody  Jan 20, 2019
#16
-2

Melody, the asker is not responsible for my rudity

Guest Jan 20, 2019
#24
+533
+1

asdf335  Jan 20, 2019
#31
+701
+1

Vieta's formulas.

Say we have a polynomial $$ax^2 + bx + c = 0$$ with solutions X_1 and X_2.

Vieta's formulas prove that...

a) X_1 + X_2 = -b/a.

b) (X_1) * (X_2) = c/a.

- PM

PartialMathematician  Jan 27, 2019
#4
+533
-1

that means there is more, because in that case, we are correct

Jan 19, 2019
#5
-3

Wrong again.

Guest Jan 19, 2019
#6
+533
0

Jan 19, 2019
#7
-3

Giving you the answer? where's the fun in that?

besides, you are downvoting my posts and that is very rude

Guest Jan 19, 2019
#32
+701
+2

You are a guest, so it does not matter. If you have a real accout, then downvoting is truly dissappointing. (*hint hint* i am continuously donvoted by someone)

PartialMathematician  Jan 27, 2019
#8
+533
0

o.o

because you are saying that us two are wrong when we are right.

unless you mean when c=c^2-2

Jan 19, 2019
#9
+533
0

now pls tell me the answer so i can work backwards

Jan 19, 2019
#10
-2

I won't give you the answer, but I will give you a hint.

Guest Jan 19, 2019
#11
+533
0

sure!

Jan 19, 2019
#12
-2

The hint is: read the question again.

Good luck ;)

Guest Jan 19, 2019
#13
+533
0

reading it again, i still dont get it o.o

Jan 19, 2019
#17
+1684
+4

I suspect this rude, careless dumb-dumb intended this equation:

$$\LARGE x^2 - ax + b = 0\\$$

Geometric solutions for this quadratic form are found by plotting A(0, 1) and B(a, b) and using the midpoint and distance formulas for finding (y) intercepts in circles. This seems consistent with the question.  Examples of these are sometimes included in pre-calculus texts.

The first use of this form comes from Thomas Carlyle (1795–1881), known more for satirical social commentary than mathematics.

GA

Jan 20, 2019
edited by GingerAle  Jan 20, 2019
#18
-1

Guest Jan 20, 2019
#19
+1684
+2

How many times do you want us to read this goddamn question?

Speak your piece, or forever STFU!

GA

GingerAle  Jan 20, 2019
#22
+533
+2

you must be having a good laugh with your retarded SMILEY FACES, ARENT YOU!?!

asdf335  Jan 20, 2019
#25
0

Sorry, I just want to see cphill's answer. I want to see his solution.

Guest Jan 20, 2019
#20
+533
0

Vieta's formulas are formulas that are like shortcuts.

two of them are where you have ax^2+bx+c.

and the sum of roots is -b/a

product is c/a.

Jan 20, 2019
#21
+533
0

vietas is also easy to prove.

lets say you have x^2+bx+c

a=1 for simplicity, it will work for any other case too

so you have (x+j)(x+k).

x^2+(j+k)x+jk.

j and k are the opposites of the roots.

so then -b/a is the sum.

and two negations still have the same product, so it is still c/a.

hope this proof helped!

Jan 20, 2019
#23
+533
+1

i downvoted all the the dumbo's posts in here.

no worries guys

Jan 20, 2019
#26
+4296
+1

Sorry. Nice explanation, asdf!

tertre  Jan 20, 2019
edited by tertre  Jan 20, 2019
edited by tertre  Jan 20, 2019
#29
+1684
0

Down voting a guest is a symbolic statement, Tertre.

In this case it’s also a catalytic condiment for a symbolic and hardy, “STFU!”

GingerAle  Jan 20, 2019
#27
+533
+1

pleez tertre give us a good explanation or something

Jan 20, 2019
#28
0

I just want to see how other users tackle this question, sorry for being rude before.

Jan 20, 2019
#30
+533
+1

so me, rom, gingerale, and tertre do not count.

ok then.. i see..

Jan 21, 2019