Solve sqrt{5-2x}-x=-1

$$\sqrt { 5 - 2 x } - x = - 1$$

Solve for x

$$x=2$$

Solution Steps

Subtract -x from both sides of the equation.

$$\sqrt{5-2x}=-1+x$$

Square both sides of the equation.

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

Calculate \sqrt{5-2x} to the power of 2 and get 5-2x.

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

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

$$5-2x=1-2x+x^{2}$$

Subtract 1 from both sides.

$$5-2x-1=-2x+x^{2}$$

Subtract 1 from 5 to get 4.

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

Add 2x to both sides.

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

Combine -2x and 2x to get 0.

$$4=x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}=4$$

Subtract 4 from both sides.

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

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

$$\sqrt{5-2\times 2}-2=-1$$

Simplify. The value x=2 satisfies the equation.

$$-1=-1$$

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

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

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$5=-1$$

Equation \sqrt{5-2x}=x-1 has a unique solution.

$$x=2$$