Solve 8(x)=x^2-2x-3/x^2-1

$$8 ( x ) = \frac { x ^ { 2 } - 2 x - 3 } { x ^ { 2 } - 1 }$$

Solve for x (complex solution)

$$x=\frac{9+\sqrt{15}i}{16}\approx 0.5625+0.242061459i$$

$$x=\frac{-\sqrt{15}i+9}{16}\approx 0.5625-0.242061459i$$

Solution Steps

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right).

$$8x\left(x-1\right)\left(x+1\right)=x^{2}-2x-3$$

Use the distributive property to multiply 8x by x-1.

$$\left(8x^{2}-8x\right)\left(x+1\right)=x^{2}-2x-3$$

Use the distributive property to multiply 8x^{2}-8x by x+1 and combine like terms.

$$8x^{3}-8x=x^{2}-2x-3$$

Subtract x^{2} from both sides.

$$8x^{3}-8x-x^{2}=-2x-3$$

Add 2x to both sides.

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

Combine -8x and 2x to get -6x.

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

Add 3 to both sides.

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

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

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

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 8. List all candidates \frac{p}{q}.

$$±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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 8x^{3}-x^{2}-6x+3 by x+1 to get 8x^{2}-9x+3. Solve the equation where the result equals to 0.

$$8x^{2}-9x+3=0$$

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 8 for a, -9 for b, and 3 for c in the quadratic formula.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 8\times 3}}{2\times 8}$$

Do the calculations.

$$x=\frac{9±\sqrt{-15}}{16}$$

Solve the equation 8x^{2}-9x+3=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{15}i+9}{16} x=\frac{9+\sqrt{15}i}{16}$$

Remove the values that the variable cannot be equal to.

$$x\in \emptyset$$

List all found solutions.

$$x=-1 x=\frac{-\sqrt{15}i+9}{16} x=\frac{9+\sqrt{15}i}{16}$$

Variable x cannot be equal to -1.

$$x=\frac{9+\sqrt{15}i}{16} x=\frac{-\sqrt{15}i+9}{16}$$