4x squared +12x=0

Samson Kaufman
2022-08-08
Answered

(4x-3)squared=4 solve the equation

4x squared +12x=0

4x squared +12x=0

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vibrerentb

Answered 2022-08-09
Author has **21** answers

For the first one:

$(4x-3{)}^{2}=4$ 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, $4{x}^{2}+12x=0$ start by factoring out anx:

$4{x}^{2}+12x=0$

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

$(4x-3{)}^{2}=4$ 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, $4{x}^{2}+12x=0$ start by factoring out anx:

$4{x}^{2}+12x=0$

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

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.

$(4x-3{)}^{2}=4$

Then you foil which multiplies all the values

(4x-3)(4x-3)=4

$16{x}^{2}-12x-12x+9=4$

$16{x}^{2}-24x+9=4$

$16{x}^{2}-24x+5=0$

(4x-5)(4x-1)=0

Then set each part equal to zero and solve for X

4x-5=0

$x=\frac{5}{4}\leftarrow $ first solution

4x-1=0

$x=\frac{1}{4}\leftarrow $ 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\leftarrow $ 1st solution

4x+12=0

$x=-\frac{12}{4}=-3\leftarrow $ 2nd solution

$(4x-3{)}^{2}=4$

Then you foil which multiplies all the values

(4x-3)(4x-3)=4

$16{x}^{2}-12x-12x+9=4$

$16{x}^{2}-24x+9=4$

$16{x}^{2}-24x+5=0$

(4x-5)(4x-1)=0

Then set each part equal to zero and solve for X

4x-5=0

$x=\frac{5}{4}\leftarrow $ first solution

4x-1=0

$x=\frac{1}{4}\leftarrow $ 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\leftarrow $ 1st solution

4x+12=0

$x=-\frac{12}{4}=-3\leftarrow $ 2nd solution

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