已知二次函数f(x)=x^2+px+q,且方程f(x)=0与f(2x)=0有相同的非零实根.(1)求q/p^2的值;(2)若f(1)=28,解方程f(x)=0

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已知二次函数f(x)=x^2+px+q,且方程f(x)=0与f(2x)=0有相同的非零实根.(1)求q/p^2的值;(2)若f(1)=28,解方程f(x)=0已知二次函数f(x)=x^2+px+q,且

已知二次函数f(x)=x^2+px+q,且方程f(x)=0与f(2x)=0有相同的非零实根.(1)求q/p^2的值;(2)若f(1)=28,解方程f(x)=0
已知二次函数f(x)=x^2+px+q,且方程f(x)=0与f(2x)=0有相同的非零实根.
(1)求q/p^2的值;
(2)若f(1)=28,解方程f(x)=0

已知二次函数f(x)=x^2+px+q,且方程f(x)=0与f(2x)=0有相同的非零实根.(1)求q/p^2的值;(2)若f(1)=28,解方程f(x)=0
1)
f(x)=0即:x²+px+q=0.1
f(2x)=0即:(2x)²+p(2x)+q=0.2
1式×4 -2式得:2px+3q=0,x=-3q/2p
2式- 1式×2得:2x²-q=0,x²=q/2
∴q/2=(-3q/2p)²
q/p²=2/9
2)
f(1)=28,即:1+p+q=28
p+q=27
又q/p²=2/9
联立两方程求p=9,q=18,或p=-27/2,q=81/2
当p=9,q=18时,f(x)=x²+9x+18
f(x)=0即:x²+9x+18=0
x1=-3,x2=-6
当p=-27/2,q=81/2时,f(x)=x²-27x/2 +81/2
f(x)=0即:x²-27x/2 +81/2=0
x1=9/2,x2=9

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good crisp blue piercing brightly color
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