What two integers does \sqrt21 fall between?
1 answer:
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= > x² + 7x + 12 = 12
= > x² + ( 4 + 3 )x + 12 = 12
= > x² + 4x + 3x + 12 = 12
= > x( x + 4 ) + 3( x + 4 ) = 12
= > ( x + 4 ) ( x + 3 ) = 12
Percy did correct till this step. But by doing like this, Percy can't get the values of the variable x.
Percy should follow the following steps :
= > x² + 7x + 12 = 12
Add -12 on both sides,
= > x² + 7x + 12 - 12 = 12 - 12
= > x² + 7x = 0
= > x( x + 7 ) = 0
= > ( x = 0 ) or ( x + 7 = 0 )
= > ( x = 0 ) or ( x = - 7 )
Hence, required value(s) of x is 0 or -7
= > x² + ( 4 + 3 )x + 12 = 12
= > x² + 4x + 3x + 12 = 12
= > x( x + 4 ) + 3( x + 4 ) = 12
= > ( x + 4 ) ( x + 3 ) = 12
Percy did correct till this step. But by doing like this, Percy can't get the values of the variable x.
Percy should follow the following steps :
= > x² + 7x + 12 = 12
Add -12 on both sides,
= > x² + 7x + 12 - 12 = 12 - 12
= > x² + 7x = 0
= > x( x + 7 ) = 0
= > ( x = 0 ) or ( x + 7 = 0 )
= > ( x = 0 ) or ( x = - 7 )
Hence, required value(s) of x is 0 or -7