Quadratic Problem: Find Sum of Roots of 4 Equations

[erisedk]

Homework Statement

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

Homework Equations

The Attempt at a Solution

##a+b=10c##
##c+d=10a##
## ab=-11d##
##cd=-11b##

Four equations, four unknowns. Obviously, this isn't supposed to be solved using regular elimination. It gets way too terrible. I can't think of a better way though. Please help.

 

[member 587159]

10(a + c)

 

[ehild]

Homework Statement

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

Homework Equations

The Attempt at a Solution

##a+b=10c##
##c+d=10a##
## ab=-11d##
##cd=-11b##

Four equations, four unknowns. Obviously, this isn't supposed to be solved using regular elimination. It gets way too terrible. I can't think of a better way though. Please help.

@Math_QED is right, the sum is 10(a+c). So isolate b and d from the first two equations and substitute into the third and fourth. See what you get for a+c.

 

[erisedk]

Thank you!
b = 10c - a
d = 10a - c
a(10c - a) = -11d
c(10a - c) = -11b
c² - a² = 11(b - d)
a + c = 121
b + d = 9 (a + c)
So a + b + c + d = 1210

 

[ehild]

Thank you!
b = 10c - a
d = 10a - c
a(10c - a) = -11d
c(10a - c) = -11b
c² - a² = 11(b - d)

You can divide the equation with c-a as a,b,c,d are all different numbers.

a + c = 121
b + d = 9 (a + c)
So a + b + c + d = 1210

Well done!