if alpha and bita are the zeroes of ax^2+bx+c then evaluate alpha^4+bita^4

Answer 1

I will use m and n for alpha and beta

from the equation we know
m+n = -b/a
mn = c/a

(m+n)^2 = m^2 + n^2 + 2mn
so m^2 + n^2 = (m+n)^2 - 2mn

also
(m^2 + n^2)^2 = m^4 + n^4 + 2m^2n^2

so m^4 + n^4 = (m^2 + n^2)^2 - 2(mn)^2
=[(m+n)^2 - 2mn]^2 - 2(mn)^2
= [(-b/a)^2 - 2c/a]^2 - 2c^2/a^2
= [b^2/a^2 - 2c/a]^2 - 2c^2/a^2

let's test it with x^2 - 7x + 12 = 0
roots m and n , where m = 4, n=3
m+n= 7
mn = 12
according to my answer
m^4 + n^4 = [(49/1-2(12)]^2 - 2(144))/1 = 25^2 - 288 = 337

actual m^4 +n^4 = 4^4 + 3^4 = 256 + 81 = 337

YEAH!

Answer 2

For ax^2+bx+c=0
the zeroes are:
α,β = (-b±sqrt(b²-4ac))/2a
=(p±q)
where
p=-b/2a
q=sqrt(b²-4ac)/2a

So to evaluate
α^4+β^4
=(p+q)^4+(p-q)^4
=2(p^4+6p²q²+q^4)
=2[(-b/2a)^4+6(-b/2a)²(sqrt(b²-4ac)/2a)²+(sqrt(b²-4ac)/2a)^4]

Expand and simplify to get:
2c²/a²-4b²c/a³+(b/a)^4

Check me.

Answer 3

If α and β are zero of f(x)=ax
2
+bx+c. The evaluate α
4
×β
4

Note that we have
=(α+β)
4
=α
4
+β
4
+
4
C
1
​
α
3
β
1
+
4
C
2
​
α
2
β
2
+
4
C
3
​
α
1
β
1
3
Then, α
4
+β
4
=(α+β)
4
−4(α
3
β
1
+α
1
β
3
)−6α
2
β
2

=(α+β)
4
−4αβ(α
2
+β
2
)−6(αβ)
2

Maxover, we have
α+β=−
a
b
​
and αβ=
a
c
​

Then,
α
4
+β
4
=(−
a
b
​
)
4
−4(
a
c
​
)[(−
a
b
​
)
2
−2(
a
c
​
)]−6(
a
c
​
)
2

=
a
4

b
4

​
−
a
4c
​
(
a
2

b
2

​
)+8(
a
c
​
)
2
−6(
a
c
​
)
2

=
a
4

b
4

​
−
a
3

4b
2
c
​
+2
a
2

C
2

​