Asked by sp
solve the following equations:
y^2-1/4y=0
y^2-1/4y=0
Answers
Answered by
Bosnian
If y^2-1/4y=0 mean :
y ^ 2 - ( 1 / 4 ) y = 0 then :
Write y ^ 2 like :
4 y ^ 2 / 4
Now :
4 y ^ 2 / 4 - ( 1 / 4 ) y = 0
( 1 / 4 ) 4 y - ( 1 / 4 ) y = 0
( 1 / 4 ) ( 4 y ^ 2 - y ) = 0
( 1 / 4 ) y ( 4 y - 1 ) = 0
Now you must solwe two equations :
( 1 / 4 ) y = 0
and
4 y - 1 = 0
( 1 / 4 ) y = 0 Multiply both sides by 4
y = 0
4 y - 1 = 0 Add 1 to both sides
4 y - 1 + 1 = 0 + 1
4 y = 1 Divide both sides by 4
y = 1 / 4
The solutions are :
y = 0 and y = 1 / 4
y ^ 2 - ( 1 / 4 ) y = 0 then :
Write y ^ 2 like :
4 y ^ 2 / 4
Now :
4 y ^ 2 / 4 - ( 1 / 4 ) y = 0
( 1 / 4 ) 4 y - ( 1 / 4 ) y = 0
( 1 / 4 ) ( 4 y ^ 2 - y ) = 0
( 1 / 4 ) y ( 4 y - 1 ) = 0
Now you must solwe two equations :
( 1 / 4 ) y = 0
and
4 y - 1 = 0
( 1 / 4 ) y = 0 Multiply both sides by 4
y = 0
4 y - 1 = 0 Add 1 to both sides
4 y - 1 + 1 = 0 + 1
4 y = 1 Divide both sides by 4
y = 1 / 4
The solutions are :
y = 0 and y = 1 / 4
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