F.4 a.math²³

(a)
α+β=-a
αβ=b

(√α+√β)^2=(α+β+2√αβ)=(-a+2√b)
(√α+√β)
=√(-a+2√b)

√(αβ)
=√b

an equation whose roots are √α,√βis
x-√(-a+2√b)+√b=0
(b)
x - 43x + 16 = 0
D=(-43)-4(1)(16)
=1849-64
=1785
>0

Therefore,the equation has two real roots

y-intercept of the equation is 16>0

Therefore,the equation has two positive or two negetive real roots

axis of symmetry is x=(43/2) x=21.5

Therefore,the equation has two positive real roots
(c)
[α^(1/4)+β^(1/4)]^2
=(√α+√β)+2(αβ)^(1/4)
=√(43+8)+4
=√51+4
(αβ)^(1/4)
=2
an equation whose roots are ^4√α, ^4√β is
x^2-(4+√51)x+2=0
(d)
(5 + √17)^4 + (5 - √17)^4.
=2[5^4+6(5^2)(17)+17^2]
=6928

(a)
α+β=-a
αβ=b

√α+√β
=√(α+β+2√αβ)
=√(-a+2√b)

√(αβ)
=√b

an equation whose roots are √α,√βis
x²-√(-a+2√b)x+√b=0

2008-12-30 16:30:42 補充：
(b)
x² - 43x + 16 = 0
D=(-43)²-4(1)(16)
=1849-64
=1785
>0

Therefore,the equation has two real roots

y-intercept of the equation is 16>0

Therefore,the equation has two positive or two negetive real roots

axis of symmetry is x=(43/2) x=21.5

Therefore,the equation has two positive real roots

2008-12-30 16:30:48 補充：
(c)
from(a)
an equation whose roots are √α and √βis
x²-√(43+2√16)x+√16=0
x²-√51x+4=0

an equation whose roots are ^4√α and ^4√βis
x²-√(√51+2√4)x+√4=0
x²-√(4+√51)x+2=0

2008-12-30 16:43:22 補充：
(d)
(5 + √17)^4 + (5 - √17)^4

The equation whose roots are (5 + √17) and (5 - √17) is
x²-(5+√17+5 - √17)x+(5 + √17)(5 - √17)=0
x²-10x+8=0

Let x²+ax+b=0 be the equation whose roots are (5 + √17)^2 and (5 - √17)^2

-√(-a+2√b)=-10
√b=8

2008-12-30 16:43:31 補充：
b=64
a=-84

The equation whose roots are (5 + √17)^2 and (5 - √17)^2 is
x²-84x+64=0

Let x²+cx+d=0 be the equation whose roots are (5 + √17)^4 and (5 - √17)^4

2008-12-30 16:43:34 補充：
-√(-c+2√d)=-84
√d=64

d=4096
c=-6928

The equation whose roots are (5 + √17)^4 and (5 - √17)^4 is x²-6928x+4096=0

Therefore (5 + √17)^4+(5 - √17)^4=-(-6928)=6928