How many positive integers a less than 100 have a corresponding integer b divisible by 3 such that the roots of x^2-ax+b=0 are consecutive positive integers?

50-60=10

step-by-step explanation:

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2

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