An algebra problem by Dinesh Nath Goswami

Algebra Level 2

Solve for real x and y .

x 2 + y 2 = 6 x + 8 y − 25 { x }^{ 2 }+{ y }^{ 2 }=6x+8y-25

Cannot be determined x=5, y=0 x=8,y=-1 x=3, y =4

Dinesh Nath Goswami by Dinesh Nath Goswami , 20, India

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1 solution

The equation can be rearranged as,

x 2 − 6 x + y 2 − 8 y + 25 = 0 { x }^{ 2 }-6x+{ y }^{ 2 }-8y+25=0

or x 2 − 6 x + 9 + y 2 − 8 y + 16 = 0 { x }^{ 2 }-6x+9+{ y }^{ 2 }-8y+16=0

or ( x − 3 ) 2 + ( y − 4 ) 2 = 0 { \left( x-3 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=0

Since x and y are real, the minimum possible value of LHS is 0, because it is the sum of squares of real expressions.

It leads to the solution x=3 & y=4 .

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