Solve 4(1+1/x)=x^2-1

$$4 ( 1 + \frac { 1 } { x } ) = x ^ { 2 } - 1$$

Solve for x

$$x=\frac{1-\sqrt{17}}{2}\approx -1.561552813$$

$$x=-1$$

$$x = \frac{\sqrt{17} + 1}{2} \approx 2.561552813$$

Steps Using the Quadratic Formula

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

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

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

$$4\left(1+\frac{1}{x}\right)x=x^{3}+x\left(-1\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$4\left(\frac{x}{x}+\frac{1}{x}\right)x=x^{3}+x\left(-1\right)$$

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

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

Express 4\times \frac{x+1}{x} as a single fraction.

$$\frac{4\left(x+1\right)}{x}x=x^{3}+x\left(-1\right)$$

Express \frac{4\left(x+1\right)}{x}x as a single fraction.

$$\frac{4\left(x+1\right)x}{x}=x^{3}+x\left(-1\right)$$

Use the distributive property to multiply 4 by x+1.

$$\frac{\left(4x+4\right)x}{x}=x^{3}+x\left(-1\right)$$

Use the distributive property to multiply 4x+4 by x.

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

Subtract x^{3} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{3} times \frac{x}{x}.

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

Since \frac{4x^{2}+4x}{x} and \frac{x^{3}x}{x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 4x^{2}+4x-x^{3}x.

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

Subtract x\left(-1\right) from both sides.

$$\frac{4x^{2}+4x-x^{4}}{x}-x\left(-1\right)=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x\left(-1\right) times \frac{x}{x}.

$$\frac{4x^{2}+4x-x^{4}}{x}-\frac{x\left(-1\right)x}{x}=0$$

Since \frac{4x^{2}+4x-x^{4}}{x} and \frac{x\left(-1\right)x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4x^{2}+4x-x^{4}-x\left(-1\right)x}{x}=0$$

Do the multiplications in 4x^{2}+4x-x^{4}-x\left(-1\right)x.

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

Combine like terms in 4x^{2}+4x-x^{4}+x^{2}.

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

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

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

Substitute t for x^{2}.

$$-t^{2}+5t+4=0$$

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 -1 for a, 5 for b, and 4 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-5±\sqrt{41}}{-2}$$

Solve the equation t=\frac{-5±\sqrt{41}}{-2} when ± is plus and when ± is minus.

$$t=\frac{5-\sqrt{41}}{2} t=\frac{\sqrt{41}+5}{2}$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=\frac{\sqrt{2\sqrt{41}+10}}{2} x=-\frac{\sqrt{2\sqrt{41}+10}}{2}$$