Question

Solve x³-7x² + 36-0, given one root being twice the other.​

Answer 1

Answer:

if the roots are a,2a,b then

a+2a+b = 7

2a^2b = -36

a*2a + ab + 2ab = 0

or, a(2a+3b) = 0

so, b = -2a/3

Solve those and you find 3,6,-2

Step-by-step explanation:

Answer 2

Answer:

ahree I'm sorry I don't know the answer to Ur question...