Let c(n) be the constant term in the expansion of
(x + 9)n.
prove by induction that
c(n) = 9n
for all n is in n.
(induction on n.) the constant term of
(x + 9)1
is

= 9
.

suppose as inductive hypothesis that the constant term of
(x + 9)k − 1
is

for some
k > 1.

then
(x + 9)k = (x + 9)k − 1 ·

,
so its constant term is

· 9 =
,
as required.

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

really your just stupid if you don't know this

step-by-step explanation: