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Question

# Solve the following equations:y2(4x2−108)=x(x3−9y3),2x2+9xy+y2=108.

A
(±23,23)
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B
(±3,3)
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C
(±25,25)
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D
(2,5)
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Solution

## The correct option is A (±2√3,∓2√3)2x2+9xy+y2=108......(i)y2(4x2−108)=x(x3−9y3)substituting (i) in place of 108y2(4x2−2x2−9xy−y2)=x4−9xy32x2y2−9xy3−y4=x4−9xy3x4+y4−2x2y2=0(x2−y2)2=0x2=y2⇒y=±xsubstituting y in (i)(1) y=x2x2+9x(x)+x2=10812x2=108x2=9x=±3⇒y=±3(2) y=−x2x2+9x(−x)+x2=108−6x2=108x2=−18x=±√−18=±2√3y=−x⇒y=∓2√3So the values of x are ±3,±2√3 and corresponding values of y are ±3,∓2√3

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