the root of the equation x^2+x-p(p+1)=0 where p is a constant
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Hi ,
x² + x - p( p+1 ) = 0
x² + ( p + 1 ) x - px - p ( p + 1 ) = 0
x [ x + p + 1 ] - p [ x + p + 1 ] = 0
( x + p + 1 ) ( x - p ) = 0
Therefore ,
x + p + 1 = 0 or x - p = 0
x = - ( p + 1 ) or x = p
I hope this helps you.
: )
x² + x - p( p+1 ) = 0
x² + ( p + 1 ) x - px - p ( p + 1 ) = 0
x [ x + p + 1 ] - p [ x + p + 1 ] = 0
( x + p + 1 ) ( x - p ) = 0
Therefore ,
x + p + 1 = 0 or x - p = 0
x = - ( p + 1 ) or x = p
I hope this helps you.
: )
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