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超簡單自創數學題
112 個回應
If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2

step?:o)


雖然中學生都做到
但係睇唔出邊到簡單


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2

step?:o)

岩唔岩先得架?:o)


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2

step?:o)

岩唔岩先得架?:o)

唔答你住


Since Ax^2+Bx+C+D=0 has real root 1
ie A(1)^2+B(1)+C+D=0 (given)
discriminant>=0
B^2-4A(C+D)>=0---(1)
Similarly, because (A+B)x^2+Cx+D has real roots 1
C^2-4(A+B)D>=0---(2)
(1)+(2):
B^2+C^2-4(A(C+D)+(A+B)D)>=0
B^2+C^2-4(AC+AD+AD+BD)>=0


For all x=0, put y=0
We have f(x乘0)=f(x)f(0)
f(0)=f(x)f(0)
f(x)=1


For all x=0, put y=0
We have f(x乘0)=f(x)f(0)
f(0)=f(x)f(0)
f(x)=1

nice try#good#


Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.#bye#

f(x) = 1/x
f(y) = 1/y
f(xy) = 1/xy = f(x)f(y)
搵到一個答案 #yup#


Oh原來第一條有古惑
f(x)=x^p for some real numbers p
f(x)f(y)=(x^p)(y^p)=(xy)^p=f(xy)


f(x)=1係special case.


Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.#bye#

f(x) = 1/x
f(y) = 1/y
f(xy) = 1/xy = f(x)f(y)
搵到一個答案 #yup#


應該唔得,因為0嘅時候係undefined
不過如果幫0加exception嘅話,咁就無問題


最衰我無寫f(x) is defined for all real numbers x


Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.#bye#

f(x) = 1/x
f(y) = 1/y
f(xy) = 1/xy = f(x)f(y)
搵到一個答案 #yup#


應該唔得,因為0嘅時候係undefined
不過如果幫0加exception嘅話,咁就無問題

無錯, 因為f(0)應該undefined.


最衰我無寫f(x) is defined for all real numbers x


你題目寫for any real number x, y 其實implies咗f(x)一定要well-defined for all real number x


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.


My instinct is to find an expression for a^4+b^4 in terms of ab and a+b

a^4+b^4
=(a+b)^4 - (4a^3)(b) - (6a^2)(b^2) - (4a)(b^3)
=(a+b)^4 - 6(ab)^2 - 4(ab)(a^2+b^2)
=(a+b)^4 - 6(ab)^2 - 4(ab)((a+b)^2-2ab))

and substitute the corresponding coefficients
Although there could be a quicker way.


function 有得find all 架咩[sosad]


Oh原來第一條有古惑
f(x)=x^p for some real numbers p
f(x)f(y)=(x^p)(y^p)=(xy)^p=f(xy)

but the question says f(0) is not 0


Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.#bye#

f(x) = 1/x
f(y) = 1/y
f(xy) = 1/xy = f(x)f(y)
搵到一個答案 #yup#




If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.


My instinct is to find an expression for a^4+b^4 in terms of ab and a+b

a^4+b^4
=(a+b)^4 - (4a^3)(b) - (6a^2)(b^2) - (4a)(b^3)
=(a+b)^4 - 6(ab)^2 - 4(ab)(a^2+b^2)
=(a+b)^4 - 6(ab)^2 - 4(ab)((a+b)^2-2ab))

and substitute the corresponding coefficients
Although there could be a quicker way.

a+b同ab係咩先


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2

得B無A既???


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2

得B無A既???

https://upload.cc/i1/2020/01/12/ap5FBy.png


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2

得B無A既???

https://upload.cc/i1/2020/01/12/ap5FBy.png

我既問題係點解A會=1?


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.


My instinct is to find an expression for a^4+b^4 in terms of ab and a+b

a^4+b^4
=(a+b)^4 - (4a^3)(b) - (6a^2)(b^2) - (4a)(b^3)
=(a+b)^4 - 6(ab)^2 - 4(ab)(a^2+b^2)
=(a+b)^4 - 6(ab)^2 - 4(ab)((a+b)^2-2ab))

and substitute the corresponding coefficients
Although there could be a quicker way.

a+b同ab係咩先


OK. I will play along.
a+b = -B/A, ab = B/A

Substituting into the obtained result, yields,

a^4+b^4 = (B/A)^4 - 4(B/A)^3 + 2(B/A)^2


If a,b are roots of quadratic equation Ax^2+Bx+C+D=0
find a^4+b^4 if B=C+D.

B^4-4B^3+2B^2

得B無A既???

https://upload.cc/i1/2020/01/12/ap5FBy.png

我既問題係點解A會=1?

我諗過了
x^2+(-a-b)x+ab=0---(1)
Ax^2+Bx+C+D=0---(2)
From (2): x^2+(B/A)x+(B/A)=0---(3)
Compare (1) and (3)
B/A=-a-b=ab
結論a+b,ab inversely proportional to A。


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