If a varies jointly with b squared and c, what is the constant of variation when a=32, b=2 and c=6?

126 maybe. sorry

1

x = 1

step-by-step explanation:

0 ÷ x = 0

0 ÷ 1 = 0

i'm not really sure how to answer that

step-by-step explanation:

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0