- #1

RChristenk

- 64

- 9

- Homework Statement
- If ##a+b \varpropto a-b ##, prove that ##a^2+b^2 \varpropto ab##

- Relevant Equations
- Basic fractions and algebra

##a+b \varpropto a-b## means ##a+b = k_1(a-b)##

##(a+b)^2=k_1^2(a-b)^2##

##a^2+b^2+2ab=a^2k_1^2+b^2k_1^2-2abk_1^2##

##2ab(1+k_1^2)=a^2(k_1^2-1)+b^2(k_1^2-1)##

##2ab(k_1^2+1)=(a^2+b^2)(k_1^2-1)##

##a^2+b^2=ab(\dfrac{2k_1^2+2}{k_1^2-1})##

Since ##a^2+b^2 \varpropto ab## means ##a^2+b^2=abk_2##, compared to what I got, I can say ##k_2 = \dfrac{2k_1^2+2}{k_1^2-1}##. Therefore ##a^2+b^2 \varpropto ab## has been proven. Is this a correct? Thanks.

##(a+b)^2=k_1^2(a-b)^2##

##a^2+b^2+2ab=a^2k_1^2+b^2k_1^2-2abk_1^2##

##2ab(1+k_1^2)=a^2(k_1^2-1)+b^2(k_1^2-1)##

##2ab(k_1^2+1)=(a^2+b^2)(k_1^2-1)##

##a^2+b^2=ab(\dfrac{2k_1^2+2}{k_1^2-1})##

Since ##a^2+b^2 \varpropto ab## means ##a^2+b^2=abk_2##, compared to what I got, I can say ##k_2 = \dfrac{2k_1^2+2}{k_1^2-1}##. Therefore ##a^2+b^2 \varpropto ab## has been proven. Is this a correct? Thanks.

Last edited by a moderator: