Question

If the roots of the equation x^2-10cx - 11d=0 are a,b and those of the x^2-10ax-11b=0 are c and d
then find the value of a+b+c+d
(a,b,c,d are distinct)

Answer 1

@ganeshie8

Answer 2

Well first do u know what a distinct number us?

Answer 3

Sorry is

Answer 4

i do

Answer 5

Ok...So, now I ask do u have the slightest clue on how to solve this or no?

Answer 6

Here hopefully this helps: tell me when to slow down

Answer 7

Now i might know what im doing lol but not completely sure
solve for the values by looking at it, if x^2 - 10cx - 11d = (x + a)(x + b), then a + b = -10c, and a*b = -11d.

Answer 8

So aftwr you have done that
You can do the same with the other equation, x^2 - 10ax - 11d = (x + c)(x + d), c + d = -10a, and c*d = -11d, but with the second problem, if you divide both sides by d, then you get c = -11, so whatever other stuff is there, c = -11.

Answer 9

Ok u still with meh?

Answer 10

you misread

Answer 11

What

Answer 12

cd=11b... not cd=11d

Answer 13

Ahhhhh i see thank u for that

Answer 14

what kind of numbers are a,b,c,d ?
are they integers by any chance ?

Answer 15

I dont think so

Answer 16

distinct numbers doesn't imply integers? @ganeshie8

Answer 17

Wait divu see u said d instead of b bc its the secong equation right?

Answer 18

distinct' real numbers ?

Answer 19

nope just distinct numbers

Answer 20

Use Viete's formula
for p(x) = ax^2+bx +c
the roots r1,r2 satisfy
r1 + r2 = -b/a
r1*r2 = c/a

Answer 21

looks like someone did that already

Answer 22

Yes umm shall i keep going with the equation

Answer 23

Bc there are two equations

Answer 24

I believe right?

Answer 25

I got
a+b = 10c ,
ab=-11d,
c+d = 10a,
c*d = 11b

Answer 26

Hey hey dont go ahead of me jk lemme keep going

Answer 27

came out with a = 5.5(9 +/- sqrt(85) ), b = 110 - a, c = -11, and d = 11-10a, but i am not too sure how accurate that is

Answer 28

So yes jazy u r correct if u want to do it the not so fun way jk

Answer 29

your equation has an issue
`x^2 - 10cx - 11d = (x + a)(x + b)`
should be
x^2 - 10cx - 11d = (x - a)(x - b)

Answer 30

since a,b are the roots of that quadratic

Answer 31

Ohhh....

Answer 32

Yes well i dont know about u but im tired from all this math. Ur right jazy i think idk maybe im wrong

Answer 33

x^2 - 10cx - 11d = (x - a)(x - b)
x^2 - 10cx -11d = x^2 -(a+b)x + ab

equating coefficients
-10c = -(a+b)
10c = a+b

Answer 34

Maybe i did my math wrong

Answer 35

So am i correct

Answer 36

what did you get

Answer 37

Here i got a new answer maybe i did my cals again:
x2-10ax-11d = (x-c)(x-d)
-10a = -c-d => c+d = 10a
-11d = cd => -11 = c

x2-10cx-11d = (x-a)(x-b)
-10c = -b-a => a+b = 10c
-11d = ab
-11(10a-c) = a(10c-a)
-110a + 11c = 10ac - a^2
-110a + 11(-11) = 10a(-11) - a^2
a^2 = 121
a = 11, -11

Answer 38

My brain is fried literally. Who's with me?

Answer 39

oops i have an error too
a+b = 10c ,
ab=-11d,
c+d = 10a,
c*d = `-11b`

Answer 40

Ohhh so am i correct

Answer 41

you again made that same mistake

Answer 42

What

Answer 43

What grade is this math in????

Answer 44

x2-10ax-11d = (x-c)(x-d)
its not the eqn

x2-10ax-11b = (x-c)(x-d) this is

Answer 45

So im correct im pretty sure

Answer 46

you are getting confused between 'b' and 'd'

Answer 47

Ohh

Answer 48

What grade is this

Answer 49

it should be simple, but with a trick and i am not getting it

Answer 50

Yea me either what grade us this lol for the third time jk

Answer 51

No but seriously

Answer 52

wolfram has all the solutions if you want

Answer 53

don't click if you want to keep working on it
http://www.wolframalpha.com/input/?i=solve+a%2Bb+%3D+10c+%2C+ab%3D-11d%2C+c%2Bd+%3D+10a%2C+c*d+%3D+-11b

Answer 54

Ohhh i see i say google it lol

Answer 55

I should have just done this in the first place duh

Answer 56

a = -11d/b
d = -11b/c
substitute
a = (-11 (-11b/c) ) / b = 121 / c

Answer 57

So I guess that problem is solved. Right?

Answer 58

a,b are roots of x^2-10cx - 11d=0
c,d are roots of x^2-10ax-11b=0

from vieta formulas we have
\(a+b=10c\tag{1}\)
\(c+d=10a\tag{2}\)
\(\implies (a+b)-(c+d) = 10(c-a) \implies b-d = 11(c-a)\tag{3}\)
\(a^2-10ac-11d=0\tag{4}\)
\(c^2-10ac-11b=0\tag{5}\)

\((5)-(4) \)
\(c^2-a^2 = 11(b-d) = 11*11(c-a)\implies c+a = 121\tag{6}\)


from \((1), (2), (6)\) :
\(a+b+c+d = 10(c+a)=10(121)\)

Answer 59

And where was this when we needes it. @ganeshie8 jk but seriously this could have been over along time ago

Answer 60

are you in a rush to go somewhere

Answer 61

So every one did good. Lol.no im just saying my brain hurts

Answer 62

here i found the solution
https://s3.amazonaws.com/upload.screenshot.co/6853f47c79

Answer 63

@ganeshie8 thanks
yep its from a IIT archieve

Answer 64

Hope we helped @divu.mkr

Answer 65

yep thanks guys :)

Answer 66

Or confuesed u lol

Answer 67

Np

Answer 68

it seems we just need c and a are distinct

Answer 69

yep

Answer 70

Yup er doodle0

Answer 71

i am curious how wolfram got this solution using these 4 equations nice challenge problem
http://www.wolframalpha.com/input/?i=solve+a%2Bb+%3D+10*c+%2C+a*b%3D-11d%2C+c%2Bd+%3D+10a%2C+c*d+%3D+-11b%2C+find+a%2Bb%2Bc%2Bd

Answer 72

Hey hey were done with tjis question no more lol jk but seriously

Answer 73

:)

Answer 74

a+b+c+d = -220 corresponds to the case when a = c

http://www.wolframalpha.com/input/?i=solve+a%2Bb+%3D+10*c+%2C+a*b%3D-11d%2C+c%2Bd+%3D+10a%2C+c*d+%3D+-11b%2C+a%3D+c%2C+find+a%2Bb%2Bc%2Bd