SOLUTION: (rs) radical sign (rs)4x+12 +4=0

Question 59685: (rs) radical sign
(rs)4x+12 +4=0
Answer by chitra(359) (Show Source): You can put this solution on YOUR website!
The question is quite confusing because we need to know whether the radical is there for the entire expression that is,
rs(4x + 12 + 4) or only for 4x + 12.

I will solve them for you in both the ways.

1st METHOD

RS(4x+ 12 + 4) = 0
Taking square root on both sides of the above equation, we get:
4x + 12 + 4 = 0
==> 4x + 16 = 0
Subtracting 16 from both sides of the above equation, we get:
==> 4x = - 16
==> x = - 16/4
==> x = -4

2nd METHOD

RS(4x + 12) + 4 = 0
Subtracting 4 from both sides of the above equation, we get:
==> RS(4x + 12) = -4
Now, Squaring both sides of the above equation, we get:
==> (4x + 12) = (-4)* (-4)
==> (4x + 12) = 16
Now, Subtracting 12 on both sides of the above equation, we get:
==> 4x = 16 - 12
==> 4x = 4
==> x = 4/4
==> x = 1
Hence the solution.

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