Solve sqrt{3x^2+2}=2x+1

$$\sqrt { 3 x ^ { 2 } + 2 } = 2 x + 1$$

Solve for x

$$x=\sqrt{5}-2\approx 0.236067977$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x^{2}+2}\right)^{2}=\left(2x+1\right)^{2}$$

Calculate \sqrt{3x^{2}+2} to the power of 2 and get 3x^{2}+2.

$$3x^{2}+2=\left(2x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$3x^{2}+2=4x^{2}+4x+1$$

Subtract 4x^{2} from both sides.

$$3x^{2}+2-4x^{2}=4x+1$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

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

Subtract 4x from both sides.

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

Subtract 1 from both sides.

$$-x^{2}+2-4x-1=0$$

Subtract 1 from 2 to get 1.

$$-x^{2}+1-4x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-4x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4}}{2\left(-1\right)}$$

Add 16 to 4.

$$x=\frac{-\left(-4\right)±\sqrt{20}}{2\left(-1\right)}$$

Take the square root of 20.

$$x=\frac{-\left(-4\right)±2\sqrt{5}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{5}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{5}}{-2} when ± is plus. Add 4 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+4}{-2}$$

Divide 4+2\sqrt{5} by -2.

$$x=-\left(\sqrt{5}+2\right)$$

Now solve the equation x=\frac{4±2\sqrt{5}}{-2} when ± is minus. Subtract 2\sqrt{5} from 4.

$$x=\frac{4-2\sqrt{5}}{-2}$$

Divide 4-2\sqrt{5} by -2.

$$x=\sqrt{5}-2$$

The equation is now solved.

$$x=-\left(\sqrt{5}+2\right) x=\sqrt{5}-2$$

Substitute -\left(\sqrt{5}+2\right) for x in the equation \sqrt{3x^{2}+2}=2x+1.

$$\sqrt{3\left(-\left(\sqrt{5}+2\right)\right)^{2}+2}=2\left(-\left(\sqrt{5}+2\right)\right)+1$$

Simplify. The value x=-\left(\sqrt{5}+2\right) does not satisfy the equation because the left and the right hand side have opposite signs.

$$2\times 5^{\frac{1}{2}}+3=-2\times 5^{\frac{1}{2}}-3$$

Substitute \sqrt{5}-2 for x in the equation \sqrt{3x^{2}+2}=2x+1.

$$\sqrt{3\left(\sqrt{5}-2\right)^{2}+2}=2\left(\sqrt{5}-2\right)+1$$

Simplify. The value x=\sqrt{5}-2 satisfies the equation.

$$2\times 5^{\frac{1}{2}}-3=2\times 5^{\frac{1}{2}}-3$$

Equation \sqrt{3x^{2}+2}=2x+1 has a unique solution.

$$x=\sqrt{5}-2$$