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!