Solve x=x-3/x^2+6x+9div(1-6/x+3)

$$x = \frac { x - 3 } { x ^ { 2 } + 6 x + 9 } \div ( 1 - \frac { 6 } { x + 3 } )$$

Solve for x

$$x=\frac{\sqrt{13}-3}{2}\approx 0.302775638$$

$$x=\frac{-\sqrt{13}-3}{2}\approx -3.302775638$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+3}{x+3}.

$$x=\frac{\frac{x-3}{x^{2}+6x+9}}{\frac{x+3}{x+3}-\frac{6}{x+3}}$$

Since \frac{x+3}{x+3} and \frac{6}{x+3} have the same denominator, subtract them by subtracting their numerators.

$$x=\frac{\frac{x-3}{x^{2}+6x+9}}{\frac{x+3-6}{x+3}}$$

Combine like terms in x+3-6.

$$x=\frac{\frac{x-3}{x^{2}+6x+9}}{\frac{x-3}{x+3}}$$

Variable x cannot be equal to -3 since division by zero is not defined. Divide \frac{x-3}{x^{2}+6x+9} by \frac{x-3}{x+3} by multiplying \frac{x-3}{x^{2}+6x+9} by the reciprocal of \frac{x-3}{x+3}.

$$x=\frac{\left(x-3\right)\left(x+3\right)}{\left(x^{2}+6x+9\right)\left(x-3\right)}$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$x=\frac{x^{2}-9}{\left(x^{2}+6x+9\right)\left(x-3\right)}$$

Use the distributive property to multiply x^{2}+6x+9 by x-3 and combine like terms.

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

Subtract \frac{x^{2}-9}{x^{3}+3x^{2}-9x-27} from both sides.

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

Factor x^{3}+3x^{2}-9x-27.

$$x-\frac{x^{2}-9}{\left(x-3\right)\left(x+3\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{\left(x-3\right)\left(x+3\right)^{2}}{\left(x-3\right)\left(x+3\right)^{2}}.

$$\frac{x\left(x-3\right)\left(x+3\right)^{2}}{\left(x-3\right)\left(x+3\right)^{2}}-\frac{x^{2}-9}{\left(x-3\right)\left(x+3\right)^{2}}=0$$

Since \frac{x\left(x-3\right)\left(x+3\right)^{2}}{\left(x-3\right)\left(x+3\right)^{2}} and \frac{x^{2}-9}{\left(x-3\right)\left(x+3\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-3\right)\left(x+3\right)^{2}-\left(x^{2}-9\right)}{\left(x-3\right)\left(x+3\right)^{2}}=0$$

Do the multiplications in x\left(x-3\right)\left(x+3\right)^{2}-\left(x^{2}-9\right).

$$\frac{x^{4}+6x^{3}+9x^{2}-3x^{3}-18x^{2}-27x-x^{2}+9}{\left(x-3\right)\left(x+3\right)^{2}}=0$$

Combine like terms in x^{4}+6x^{3}+9x^{2}-3x^{3}-18x^{2}-27x-x^{2}+9.

$$\frac{x^{4}+3x^{3}-10x^{2}-27x+9}{\left(x-3\right)\left(x+3\right)^{2}}=0$$

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right)^{2}.

$$x^{4}+3x^{3}-10x^{2}-27x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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^{4}+3x^{3}-10x^{2}-27x+9 by x-3 to get x^{3}+6x^{2}+8x-3. Solve the equation where the result equals to 0.

$$x^{3}+6x^{2}+8x-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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}+6x^{2}+8x-3 by x+3 to get x^{2}+3x-1. Solve the equation where the result equals to 0.

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

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 1 for a, 3 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-3±\sqrt{13}}{2}$$

Solve the equation x^{2}+3x-1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{13}-3}{2} x=\frac{\sqrt{13}-3}{2}$$

Remove the values that the variable cannot be equal to.

$$x\in \emptyset$$

List all found solutions.

$$x=3 x=-3 x=\frac{-\sqrt{13}-3}{2} x=\frac{\sqrt{13}-3}{2}$$

Variable x cannot be equal to any of the values 3,-3.

$$x=\frac{\sqrt{13}-3}{2} x=\frac{-\sqrt{13}-3}{2}$$