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algebra problem plz help
- Thread starter Debasish
- Start date
Hello, Debasish!
We have: .\(\displaystyle x^2+2x+1 + y^2 \:=\:0 \quad\Rightarrow\quad (x+1)^2 + y^2 \:=\:0\)
. . Hence: .\(\displaystyle x = \text{-}1,\:y = 0\)
Therefore: .\(\displaystyle x^{31} + y^{35} \;=\;(\text{-}1)^{31} + (0)^{35} \;=\;-1 + 0 \;=\;-1\)
\(\displaystyle \text{If }x^2+y^2+2x+1\,=\,0\)
. . \(\displaystyle \text{find the value of: }\:x^{31}+y^{35}\)
We have: .\(\displaystyle x^2+2x+1 + y^2 \:=\:0 \quad\Rightarrow\quad (x+1)^2 + y^2 \:=\:0\)
. . Hence: .\(\displaystyle x = \text{-}1,\:y = 0\)
Therefore: .\(\displaystyle x^{31} + y^{35} \;=\;(\text{-}1)^{31} + (0)^{35} \;=\;-1 + 0 \;=\;-1\)
D
Deleted member 4993
Guest
I want to add a little explanation to Soroban's elegant solution.
(x+1)2 and y2 will be always positive - because those are square numbers.
Sum of two positive numbers can be equal to zero - iff each number is equal zero.
Then rest of the story....
(x+1)2 and y2 will be always positive - because those are square numbers.
Sum of two positive numbers can be equal to zero - iff each number is equal zero.
Then rest of the story....