Solve sqrt{3x^2-2}=(2x-1)

$$\sqrt { 3 x ^ { 2 } - 2 } = ( 2 x - 1 )$$

Solve for x

$$x=3$$

$$x=1$$

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}.

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

Add 4x to both sides.

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

Subtract 1 from both sides.

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

Subtract 1 from -2 to get -3.

$$-x^{2}-3+4x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+4x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

Rewrite -x^{2}+4x-3 as \left(-x^{2}+3x\right)+\left(x-3\right).

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

Factor out -x in -x^{2}+3x.

$$-x\left(x-3\right)+x-3$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+1\right)$$

To find equation solutions, solve x-3=0 and -x+1=0.

$$x=3 x=1$$

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

$$\sqrt{3\times 3^{2}-2}=2\times 3-1$$

Simplify. The value x=3 satisfies the equation.

$$5=5$$

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

$$\sqrt{3\times 1^{2}-2}=2\times 1-1$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{3x^{2}-2}=2x-1.

$$x=3 x=1$$