How do you solve #sqrt(4y+21)=y#?

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Answer 1 · 2015-10-15T09:34:18

#y# = #7#

#sqrt(4y+21)# = #y#

squaring on both sides

#(sqrt(4y+21))^2# =#y^2#
#4y +21 = y^2#

rearranging the equation

# y^2-4y -21 = 0#

# y^2-7y+3y -21 = 0#
# y(y-7)+3(y-7) = 0#
# (y-7)(y+3) = 0#
#y-7 =0 or y+3 = 0#
#y=7 and y= -3 #

Then we have to check if both these values of #y# will satisfy the original equation or not

1: When #y=7#,

#sqrt(4xx7+21)# = #7#
#or,sqrt(28+21)# = #7#
#or,sqrt(49)# = #7#
#or,7=7#

which is true, #therefore y=7#

2:When #y=-3#,
Case-1
#sqrt(4xx(-3)+21)# = #-3#
#or,sqrt(-12+21)# = #-3#
#or,sqrt(9)# = #-3#
#or,3!=-3#

#therefore y!=-3#

Therefore, the value of #y# is #7#
Case-2
#sqrt(4xx(-3)+21)# = #-3#
#or,sqrt(-12+21)# = #-3#
#or,sqrt(9)# = #-3#
#or, sqrt (-3^2)=#-3#
-3 = -3

Hence 7 and -3 are roots for this equation

Answer 2 · 2015-10-17T02:08:26

#y = 7 and -3#

#sqrt(4y+21) = y#

When #y #= -3

#sqrt(4X(-3)+21) = -3#
#sqrt(-12+21)=-3#
#sqrt(9)=-3#

case -1

#sqrt(-3^2)= -3#

-3 = -3

=0
Hence -3 also satisfied the equation, hence -3 is also root for the above equation along with 7