Solve x^2-2/x+3=9-2/x+3

$${ x }^{ 2 } - \frac{ 2 }{ x+3 } = 9- \frac{ 2 }{ x+3 }$$

Solve for x

$$x=3$$

Solution Steps

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by x+3.

$$\left(x+3\right)x^{2}-2=\left(x+3\right)\times 9-2$$

Use the distributive property to multiply x+3 by x^{2}.

$$x^{3}+3x^{2}-2=\left(x+3\right)\times 9-2$$

Use the distributive property to multiply x+3 by 9.

$$x^{3}+3x^{2}-2=9x+27-2$$

Subtract 2 from 27 to get 25.

$$x^{3}+3x^{2}-2=9x+25$$

Subtract 9x from both sides.

$$x^{3}+3x^{2}-2-9x=25$$

Subtract 25 from both sides.

$$x^{3}+3x^{2}-2-9x-25=0$$

Subtract 25 from -2 to get -27.

$$x^{3}+3x^{2}-27-9x=0$$

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

$$x^{3}+3x^{2}-9x-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}-9x-27 by x-3 to get x^{2}+6x+9. Solve the equation where the result equals to 0.

$$x^{2}+6x+9=0$$

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, 6 for b, and 9 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-6±0}{2}$$

Solutions are the same.

$$x=-3$$

Remove the values that the variable cannot be equal to.

$$x=3$$

List all found solutions.

$$x=3 x=-3$$

Variable x cannot be equal to -3.

$$x=3$$