Solve sqrt{x}=1/2sqrt{x}

$$\sqrt { x } = \frac { 1 } { 2 \sqrt { x } }$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

Square both sides of the equation.

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

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

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

To raise \frac{1}{2\sqrt{x}} to a power, raise both numerator and denominator to the power and then divide.

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

Calculate 1 to the power of 2 and get 1.

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

Expand \left(2\sqrt{x}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

$$x=\frac{1}{4x}$$

Subtract \frac{1}{4x} from both sides.

$$x-\frac{1}{4x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{4x}{4x}.

$$\frac{x\times 4x}{4x}-\frac{1}{4x}=0$$

Since \frac{x\times 4x}{4x} and \frac{1}{4x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 4x-1}{4x}=0$$

Do the multiplications in x\times 4x-1.

$$\frac{4x^{2}-1}{4x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4x.

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

Consider 4x^{2}-1. Rewrite 4x^{2}-1 as \left(2x\right)^{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).

$$\left(2x-1\right)\left(2x+1\right)=0$$

To find equation solutions, solve 2x-1=0 and 2x+1=0.

$$x=\frac{1}{2} x=-\frac{1}{2}$$

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

$$\sqrt{\frac{1}{2}}=\frac{1}{2\sqrt{\frac{1}{2}}}$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$\frac{1}{2}\times 2^{\frac{1}{2}}=\frac{1}{2}\times 2^{\frac{1}{2}}$$

Substitute -\frac{1}{2} for x in the equation \sqrt{x}=\frac{1}{2\sqrt{x}}. The expression \sqrt{-\frac{1}{2}} is undefined because the radicand cannot be negative.

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

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

$$x=\frac{1}{2}$$