Solve x^2-7=6/x

$$x ^ { 2 } - 7 = \frac { 6 } { x }$$

Solve for x

$$x=-1$$

$$x=-2$$

$$x=3$$

Solution Steps

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

$$xx^{2}+x\left(-7\right)=6$$

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

$$x^{3}+x\left(-7\right)=6$$

Subtract 6 from both sides.

$$x^{3}+x\left(-7\right)-6=0$$

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

$$x^{3}-7x-6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x-6 by x+1 to get x^{2}-x-6. Solve the equation where the result equals to 0.

$$x^{2}-x-6=0$$

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, -1 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$x=\frac{1±5}{2}$$

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

$$x=-2 x=3$$

List all found solutions.

$$x=-1 x=-2 x=3$$