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

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

Solve for x (complex solution)

$$x=-1$$

$$x=1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}+x-1} to the power of 2 and get x^{2}+x-1.

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

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

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

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

$$x=1 x=-1$$

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

$$\sqrt{1^{2}+1-1}=\sqrt{1}$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

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

Simplify. The value x=-1 satisfies the equation.

$$i=i$$

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

$$x=1 x=-1$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}+x-1} to the power of 2 and get x^{2}+x-1.

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

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

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

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

$$x=1 x=-1$$

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

$$\sqrt{1^{2}+1-1}=\sqrt{1}$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

Substitute -1 for x in the equation \sqrt{x^{2}+x-1}=\sqrt{x}. The expression \sqrt{\left(-1\right)^{2}-1-1} is undefined because the radicand cannot be negative.

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

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

$$x=1$$