Solve 2sqrt{x}+1/x=3

$$2 \sqrt { x } + \frac { 1 } { x } = 3$$

Solve for x

$$x=1$$

Solution Steps

Subtract \frac{1}{x} from both sides of the equation.

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

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

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

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x}\right)^{2}x^{2}=\left(x\times 3-1\right)^{2}$$

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

$$4xx^{2}=\left(x\times 3-1\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$4x^{3}=\left(x\times 3-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x\times 3-1\right)^{2}.

$$4x^{3}=9x^{2}-6x+1$$

Subtract 9x^{2} from both sides.

$$4x^{3}-9x^{2}=-6x+1$$

Add 6x to both sides.

$$4x^{3}-9x^{2}+6x=1$$

Subtract 1 from both sides.

$$4x^{3}-9x^{2}+6x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{1}{4},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-9x^{2}+6x-1 by x-1 to get 4x^{2}-5x+1. Solve the equation where the result equals to 0.

$$4x^{2}-5x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, -5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\times 1}}{2\times 4}$$

Do the calculations.

$$x=\frac{5±3}{8}$$

Solve the equation 4x^{2}-5x+1=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

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

$$2\sqrt{1}+\frac{1}{1}=3$$

Simplify. The value x=1 satisfies the equation.

$$3=3$$

Substitute \frac{1}{4} for x in the equation 2\sqrt{x}+\frac{1}{x}=3.

$$2\sqrt{\frac{1}{4}}+\frac{1}{\frac{1}{4}}=3$$

Simplify. The value x=\frac{1}{4} does not satisfy the equation.

$$5=3$$

Equation 2\sqrt{x}x=3x-1 has a unique solution.

$$x=1$$