Solve x^2-4x+4/x^2-4=1/(x+3)

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

Solve for x

$$x=2\sqrt{2}\approx 2.828427125$$

$$x=-2\sqrt{2}\approx -2.828427125$$

Solution Steps

Variable x cannot be equal to any of the values -3,-2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right)\left(x+3\right), the least common multiple of x^{2}-4,x+3.

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

Use the distributive property to multiply x+3 by x^{2}-4x+4 and combine like terms.

$$x^{3}-x^{2}-8x+12=x^{2}-4$$

Subtract x^{2} from both sides.

$$x^{3}-x^{2}-8x+12-x^{2}=-4$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$x^{3}-2x^{2}-8x+12=-4$$

Add 4 to both sides.

$$x^{3}-2x^{2}-8x+12+4=0$$

Add 12 and 4 to get 16.

$$x^{3}-2x^{2}-8x+16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±16,±8,±4,±2,±1$$

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}-2x^{2}-8x+16 by x-2 to get x^{2}-8. Solve the equation where the result equals to 0.

$$x^{2}-8=0$$

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, 0 for b, and -8 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$x=\frac{0±4\sqrt{2}}{2}$$

Solve the equation x^{2}-8=0 when ± is plus and when ± is minus.

$$x=-2\sqrt{2} x=2\sqrt{2}$$

Remove the values that the variable cannot be equal to.

$$x\in \emptyset$$

List all found solutions.

$$x=2 x=-2\sqrt{2} x=2\sqrt{2}$$

Variable x cannot be equal to 2.

$$x=2\sqrt{2} x=-2\sqrt{2}$$