It is known that the coefficient b and c in the equation x^2+bx+c=0 are also the roots of that equation (c≠0).

Question

Find b and c.

Answer 1

If Ax² + Bx + C = 0, then sum of roots = -B/A and product of roots = C/A

 

So,  for x²+bx+c = 0, we  have -b/(1) = b + c  and c/1 = bc

 

We are given that c≠0, so since c = bc, we have b = 1.

 

   Since -b = b+c and b = 1, we get -2b = c.  So, c = -2.

 

The equation is x² + x - 2 = 0

 

Answer 2

(-B + sqrt ( b^2 - 4c) )/2 = B

-b + sqrt ( b^2 - 4c) = 2b

sqrt( b^2 - 4c) = 3b

b^2 - 4c = 9b^2

-4c = 8b^2

c = -2b^2 <--- please label this equation ALPHA; we will need it later

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(-B - sqrt ( b^2 - 4c) )/2 = c

-b - sqrt ( b^2 - 4c) = 2c

- sqrt( b^2 - 4c) = 2c + b

b^2 - 4c = (2c + b)^2 = 4c^2 + 4bc + b^2

-4c = 4c^2 + 4bc

-c = c^2 + bc

2b^2 = (-2b^2)^2 + b(-2b^2)

2b^2 = 4b^4 - 2b^3

0 = 4b^4 - 2b^3 - 2b^2

0 = 2b^2 ( 2b^2 - b - 1)

0 = 2b^2 ( 2b + 1)( b - 1 )

b=0, -1, 1

Recall c = -2b^2 from equation ALPHA

For b=0, c=0

THe equation is x^2 = x^2 + 0x + 0 which has roots b=0 and c=0



for b=1, c=-2.
THe equation is x^2 + x - 2 which has roots (x + 2)(x - 1)=0
x=-2 = c and x = 1 = b

for b=-1, c = -2.
The equation is x^2 - x - 2 which has roots ( x - 2)(x + 1 )=0
x=2 and x=-1 = b
So the third case does not work.