Quadrinomial Equations (Problem 1, Version 3)
( x − 5 ) 2 − ( y + 5 ) 2 = 0
( x − 5 ) + ( y + 5 ) 2 = 1 2
√ ( x − 5 ) + ( y + 5 ) = 2 i + 4
√ ( x − 5 ) + √ ( y + 5 ) = 2 i + 2
Find the values of x and y
Give your answer as x + y
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The answer is 0.
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2 solutions
√ ( x − 5 ) + ( y + 5 ) = 2 i + 4
Since √ ( x − 5 ) = 2 i , x − 5 = − 4 , x = 1
Now since y + 5 = 4 , y = − 1
Now substitute 1 , − 1 into the rest of the equations:
( 1 − 5 ) 2 = 1 6 − ( − 1 + 5 ) 2 = 1 6 − 1 6 = 0 - Yes
( 1 − 5 ) = − 4 + ( − 1 + 5 ) 2 = − 4 + 1 6 = 1 6 − 4 = 1 2 - Yes
√ ( 1 − 5 ) = − 2 + ( − 1 + 5 ) = − 2 + 4 = 2 - Yes
√ ( 1 − 5 ) = − 2 + √ ( − 1 + 5 ) = − 2 + 2 = 0 - Yes
Therefore, the answer is x = 1 , y = - 1
When we subtract (4) from (3), we get:
( y + 5 ) − ( y + 5 ) ( y + 3 ) 2 y 2 + 6 y − y + 9 − 5 y = 2 = y + 5 = 0 = − 1
Substituting y in equation (2) results value of x to be:
( x − 5 ) + ( y + 5 ) 2 ( x − 5 ) + ( − 1 + 5 ) 2 x x = 1 2 = 1 2 = 1 2 − 1 6 + 5 = 1
⟹ x + y = ( − 1 ) + ( 1 ) = 0