anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
Well first do u know what a distinct number us?
anonymous
  • anonymous
Sorry is

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anonymous
  • anonymous
i do
anonymous
  • anonymous
Ok...So, now I ask do u have the slightest clue on how to solve this or no?
anonymous
  • anonymous
Here hopefully this helps: tell me when to slow down
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Ok u still with meh?
anonymous
  • anonymous
you misread
anonymous
  • anonymous
What
anonymous
  • anonymous
cd=11b... not cd=11d
anonymous
  • anonymous
Ahhhhh i see thank u for that
ganeshie8
  • ganeshie8
what kind of numbers are a,b,c,d ? are they integers by any chance ?
anonymous
  • anonymous
I dont think so
anonymous
  • anonymous
distinct numbers doesn't imply integers? @ganeshie8
anonymous
  • anonymous
Wait divu see u said d instead of b bc its the secong equation right?
ganeshie8
  • ganeshie8
distinct' real numbers ?
anonymous
  • anonymous
nope just distinct numbers
anonymous
  • anonymous
Use Viete's formula for p(x) = ax^2+bx +c the roots r1,r2 satisfy r1 + r2 = -b/a r1*r2 = c/a
anonymous
  • anonymous
looks like someone did that already
anonymous
  • anonymous
Yes umm shall i keep going with the equation
anonymous
  • anonymous
Bc there are two equations
anonymous
  • anonymous
I believe right?
anonymous
  • anonymous
I got a+b = 10c , ab=-11d, c+d = 10a, c*d = 11b
anonymous
  • anonymous
Hey hey dont go ahead of me jk lemme keep going
anonymous
  • anonymous
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
anonymous
  • anonymous
So yes jazy u r correct if u want to do it the not so fun way jk
anonymous
  • anonymous
your equation has an issue `x^2 - 10cx - 11d = (x + a)(x + b)` should be x^2 - 10cx - 11d = (x - a)(x - b)
anonymous
  • anonymous
since a,b are the roots of that quadratic
anonymous
  • anonymous
Ohhh....
anonymous
  • anonymous
Yes well i dont know about u but im tired from all this math. Ur right jazy i think idk maybe im wrong
anonymous
  • anonymous
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
anonymous
  • anonymous
Maybe i did my math wrong
anonymous
  • anonymous
So am i correct
anonymous
  • anonymous
what did you get
anonymous
  • anonymous
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
anonymous
  • anonymous
My brain is fried literally. Who's with me?
anonymous
  • anonymous
oops i have an error too a+b = 10c , ab=-11d, c+d = 10a, c*d = `-11b`
anonymous
  • anonymous
Ohhh so am i correct
anonymous
  • anonymous
you again made that same mistake
anonymous
  • anonymous
What
anonymous
  • anonymous
What grade is this math in????
anonymous
  • anonymous
x2-10ax-11d = (x-c)(x-d) its not the eqn x2-10ax-11b = (x-c)(x-d) this is
anonymous
  • anonymous
So im correct im pretty sure
anonymous
  • anonymous
you are getting confused between 'b' and 'd'
anonymous
  • anonymous
Ohh
anonymous
  • anonymous
What grade is this
anonymous
  • anonymous
it should be simple, but with a trick and i am not getting it
anonymous
  • anonymous
Yea me either what grade us this lol for the third time jk
anonymous
  • anonymous
No but seriously
anonymous
  • anonymous
wolfram has all the solutions if you want
anonymous
  • anonymous
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
anonymous
  • anonymous
Ohhh i see i say google it lol
anonymous
  • anonymous
I should have just done this in the first place duh
anonymous
  • anonymous
a = -11d/b d = -11b/c substitute a = (-11 (-11b/c) ) / b = 121 / c
anonymous
  • anonymous
So I guess that problem is solved. Right?
ganeshie8
  • ganeshie8
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 \((1), (2), (6)\) : \(a+b+c+d = 10(c+a)=10(121)\)
anonymous
  • anonymous
And where was this when we needes it. @ganeshie8 jk but seriously this could have been over along time ago
anonymous
  • anonymous
are you in a rush to go somewhere
anonymous
  • anonymous
So every one did good. Lol.no im just saying my brain hurts
ganeshie8
  • ganeshie8
here i found the solution https://s3.amazonaws.com/upload.screenshot.co/6853f47c79
anonymous
  • anonymous
@ganeshie8 thanks yep its from a IIT archieve
anonymous
  • anonymous
Hope we helped @divu.mkr
anonymous
  • anonymous
yep thanks guys :)
anonymous
  • anonymous
Or confuesed u lol
anonymous
  • anonymous
Np
ganeshie8
  • ganeshie8
it seems we just need c and a are distinct
anonymous
  • anonymous
yep
anonymous
  • anonymous
Yup er doodle0
anonymous
  • anonymous
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
anonymous
  • anonymous
Hey hey were done with tjis question no more lol jk but seriously
anonymous
  • anonymous
:)
ganeshie8
  • ganeshie8
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

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