Solve sqrt{3-2x}=x

$$\sqrt{ 3-2x } =x$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

$$3-2x=x^{2}$$

Subtract x^{2} from both sides.

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

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

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

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.

$$a+b=-2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-3$$

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

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

Factor out x in the first and 3 in the second group.

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

Factor out common term -x+1 by using distributive property.

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

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

$$x=1 x=-3$$

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

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

$$\sqrt{3-2\left(-3\right)}=-3$$

Simplify. The value x=-3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$3=-3$$

Equation \sqrt{3-2x}=x has a unique solution.

$$x=1$$