# 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**

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}.**