What is the answer to 2x-sqrt(1+x^2)=0 ?

The answer to 2x-sqrt(1+x^2)=0 is x= 1/(sqrt(3))

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2x-sqrt(1+x^2)=0

2 x - 1 + x^{2} = 0

x = \frac{1}{3}

Decimal

x = 0.57735 \dots

Solution steps

2 x - 1 + x^{2} = 0

Remove square roots

1 + x^{2} = 4 x^{2}

Solve

1 + x^{2} = 4 x^{2} : x = \frac{1}{3}, x = - \frac{1}{3}

x = \frac{1}{3}, x = - \frac{1}{3}

Verify Solutions:

x = \frac{1}{3}

True

, x = - \frac{1}{3}

False

The solution is

x = \frac{1}{3}

r, r (t) = 1 0^{2} t

- 4 = - 4 1 - 8 b

- 18 + 11 x = x

x = (x + 2)^{\frac{1}{2}}

4 x + 61 = 2 x - 1

What is the answer to 2x-sqrt(1+x^2)=0 ?
The answer to 2x-sqrt(1+x^2)=0 is x= 1/(sqrt(3))
Solve for x: 2x-sqrt(1+x^2)=0
The solution is x= 1/(sqrt(3))