(x+2) (x+3) + (x-3) (x-2)-2x(x+1)=0
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Answered by
13
We have ( x + 2 ) ( x + 3 ) + ( x - 3 ) ( x - 2 ) - 2x ( x + 1 ) = 0
So,
⇒( x2 + 3x + 2x + 6 ) +( x2 - 3x - 2x + 6 ) - ( 2x2 + 2x ) = 0
⇒( x2 + 5x + 6 ) +( x2 - 5x + 6 ) - ( 2x2 + 2x ) = 0
⇒x2 + 5x + 6 + x2 - 5x + 6 - 2x2 - 2x = 0
⇒12 - 2x = 0
⇒2x = 12
⇒x = 6 ( Ans )
So,
⇒( x2 + 3x + 2x + 6 ) +( x2 - 3x - 2x + 6 ) - ( 2x2 + 2x ) = 0
⇒( x2 + 5x + 6 ) +( x2 - 5x + 6 ) - ( 2x2 + 2x ) = 0
⇒x2 + 5x + 6 + x2 - 5x + 6 - 2x2 - 2x = 0
⇒12 - 2x = 0
⇒2x = 12
⇒x = 6 ( Ans )
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7
Hey dear here is ur answer
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