(4x-3)squared=4 solve the equation 4x squared +12x=0

Samson Kaufman 2022-08-08 Answered
(4x-3)squared=4 solve the equation
4x squared +12x=0
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Answers (2)

vibrerentb
Answered 2022-08-09 Author has 21 answers
For the first one:
( 4 x 3 ) 2 = 4 Take the square root of both sides, giving you:
4x-3=2
and
4x-3=-2
Solve both for x:
4x-3=2
4x=5
x=5/4
And
4x-3=-2
4x=1
x=1/4
So for the first one you get x=5/4 and x=1/4
For the second one, 4 x 2 + 12 x = 0 start by factoring out anx:
4 x 2 + 12 x = 0
x(4x+12)=0
From here you can say that x=0 because if you multiplied 0 bywhatever is inside the parenthesis then you would get 0. Then go onand solve for any other x's:
4x+12=0
4x=-12
x=-3
So for the second one you get x=0 and x=-3
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wendi1019gt
Answered 2022-08-10 Author has 7 answers
when you have a quadratic you have to set it equal to zero and then solve for the zeroes.
( 4 x 3 ) 2 = 4
Then you foil which multiplies all the values
(4x-3)(4x-3)=4
16 x 2 12 x 12 x + 9 = 4
16 x 2 24 x + 9 = 4
16 x 2 24 x + 5 = 0
(4x-5)(4x-1)=0
Then set each part equal to zero and solve for X
4x-5=0
x = 5 4 first solution
4x-1=0
x = 1 4 second solution
For the second question it already equals zero and because there is no C term we will just factor out an x and then we can solve
4 x 2 + 12 = 0
x(4x+12)=0
Set each part equal to zero
x = 0 1st solution
4x+12=0
x = 12 4 = 3 2nd solution
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