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If a,b,care non-zero, then the number of solutions of 2x2a2y2b2z2c2=0x2a2+2y2b2z2c2=0x2a2y2b2+2z2c2=0is

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a
6
b
9
c
8
d
Infinite

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detailed solution

Correct option is D

(2)−(3)⇒3y2b2−3z2c2=0⇒y2b2=z2c2,substitute in equation(1)(1)⇒2x2a2−2y2b2=0⇒x2a2=y2b2∴x2a2=y2b2=z2c2=k2(say)x2=a2k2,y2=b2k2,z2=c2k2⇒x=±ka,y=±kb,z=±kc,k∈R.


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Suppose  a, b,  R and a, b1. If the system of equation ax + y + z = 0x + by + z = 0  ,  x + y + 2z = 0     

has a non-trivial solution, then

 


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