## Condition on Coefficients For Coupled Quadratics To Have Solutions

Theorem
Suppose the coefficients of two quadratic equations
$x^2+b_1x+c_1=0, \: x^2+b_2x+c_2=0$
have the relationship
$b_1b_2=2(c_1+c_2)$
(1).
Then at least one of the equations have solutions. Proof Suppose neither equation has solutions. Then
$b_1^2-4c_1 , \: b_2^2-4c_2 \lt0$
.
Hence
$b_1^2-4c_1+b_2^2-4c_2 =(b_1^2+b_2^2)-4(c_1+c_2) \lt 0$
.
Use the condition (1) above to give
$(b_1^2+b_2^2)-2b_1b_2=(b_1-b_2)^2 \lt 0$
.
This is impossible since a square number is at least zero, so at least one equation has solutions.