Solve x^2+3x-4=12/x

$$x ^ { 2 } + 3 x - 4 = \frac { 12 } { x }$$

Solve for x

$$x=-2$$

$$x=2$$

$$x=-3$$

Solution Steps

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

$$xx^{2}+3xx+x\left(-4\right)=12$$

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

$$x^{3}+3xx+x\left(-4\right)=12$$

Multiply x and x to get x^{2}.

$$x^{3}+3x^{2}+x\left(-4\right)=12$$

Subtract 12 from both sides.

$$x^{3}+3x^{2}+x\left(-4\right)-12=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}+3x^{2}-4x-12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}-4x-12 by x-2 to get x^{2}+5x+6. Solve the equation where the result equals to 0.

$$x^{2}+5x+6=0$$

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 6 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{-5±1}{2}$$

Solve the equation x^{2}+5x+6=0 when ± is plus and when ± is minus.

$$x=-3 x=-2$$

List all found solutions.

$$x=2 x=-3 x=-2$$