What is x if 1-square root(13+3x^2)=2x ?
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Given the equation:
1- sqrt(13+3x^2) = 2x
We need to solve for x.
First we will move 1 to the right side.
==> - sqrt(13+3x^2) = 2x -1
Now we will square both sides.
==> (13+3x^2) = (2x-1)^2
==> 13+ 3x^2 = 4x^2 - 4x +1
Now we will combine all terms on the left side.
==> -x^2 + 4x +12 = 0
We will multiply by -1.
==> x^2 - 4x -12 = 0
Now we will factor.
==> (x-6)(x+2) = 0
==> The answer is x = 6 and x= -2
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We'll impose constraints of existence of square root:
13+3x^2 >= 0
Since it is a sum of positive amounts, we'll accept any value of x as solution of equation.
We'll subtract 2x both sides:
1 - 2x = sqrt(13+3x^2)
We'll raise to square to eliminate the square root:
1 - 4x + 4x^2 = 13+3x^2
We'll move all terms to one side and we'll cmbine like terms:
x^2 - 4x - 12 = 0
We'll apply quadratic formula:
x1 = [4+sqrt(64)]/2
x1 = 6
x2 = (4-8)/2
x2 = -2
The solutions of the equation are {-2 ; 6}.
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