指數方程問題5^(2x)-K 5^(x+2)+625=0?

5^(2x)－k*5^(x+2)+625=0，已知兩根相乘為3，求k值?
Sol
設兩根a，b，a>=b，p=5^x
ab=3
p^2－25kp+625=0，兩根5^a，5^b
5^a*5^b=625
5^(a+b)=5^4
a+b=4
b=4－a
a(4－a)=3
－a^2+4a－3=0
a^2－4a+3=0
(a－3)(a－1)=0
a=3 or a=1(不合)
b=1
5^a+5^b=25k
25k=5^3+5^1=130
k=26/5

Given:
f(x,K)=5^(2x)-K5^(x+2)+625=0
5^(2x)-K5^x*5^2+625=0
5^(2x)-25K5^x+625=0
set y=5^x
y^2-25Ky+625=0
y1(K)=(25/2)*(K+sqrt(K^2-4)), y2(K)=(25/2)*(K-sqrt(K^2-4))
therefore, solving, x=log(y)/log(5)
X1(K)=log(y1(K))/log(5), and X2(K)=log(y2(K))/log(5)
and the product,
P(K)=X1(K)*X2(K)=log(y1(K))/log(5)*log(y2(K))/log(5)
and we need to solve for P(K)=3 using Newton's method,or its variant.
With a starting estimate of K=2 (minimum possible for real values),
the successive values of K are
K=2
K=5.86265
K=5.22014
K=5.20002
K=5.20000
So K=5.2
Check:
K=5.2, x1=1, x2=3,
f(1,5.2)=0
f(3,5.2)=0
ok

Alternatively, this can be solved by trial and error for x=1,2,3...

K=5.2