Gleichung mit Parameter lösen x^2-3rx+2r^2=0

Question

Wie kann man diese gleichungen lösen?:  r,s sind parameter

 

a) x^2-3rx+2r^2=0      

b) 2x^2+(4s-r)x=2rs

 

 

Comments

x^2 - 3·r·x + 2·r^2 = 0

pq-Lösungsformel

x = - p/2 ± √((p/2)^2 - q) = - (- 3·r)/2 ± √(((- 3·r)/2)^2 - (2·r^2)) = 1.5·r ± 0.5·r

Answer 1

x² + (-3r)x + 2r² = 0

1.5r +- √(2.25r^2 - 2r^2)

1.5r +- √(0.25r^2)

1.5r +- 0.5r

x1 = r
x2 = 2r
 

Comments

2x² + (4s-r)x - 2rs = 0

x² + (4s-r)/2*x - rs = 0

-(4s-r)/4 +- √((4s-r)^2/16 + rs)

-(4s-r)/4 +- √(r^2/16 + r·s/2 + s^2)

-(4s-r)/4 +- √((r + 4·s)^2/16)

-(4s-r)/4 +- (r + 4·s)/4

x1 = r/2
x2 = -2s