Solve 2x^2/x-1+x-5/x^2-1=1

$$\frac{ 2 { x }^{ 2 } }{ x-1 } + \frac{ x-5 }{ { x }^{ 2 } -1 } = 1$$

Solve for x (complex solution)

$$x=\frac{-3+\sqrt{23}i}{4}\approx -0.75+1.198957881i$$

$$x=\frac{-\sqrt{23}i-3}{4}\approx -0.75-1.198957881i$$

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), the least common multiple of x-1,x^{2}-1.

$$\left(x+1\right)\times 2x^{2}+x-5=\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply x+1 by 2.

$$\left(2x+2\right)x^{2}+x-5=\left(x-1\right)\left(x+1\right)$$

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

$$2x^{3}+2x^{2}+x-5=\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+1\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 1.

$$2x^{3}+2x^{2}+x-5=x^{2}-1$$

Subtract x^{2} from both sides.

$$2x^{3}+2x^{2}+x-5-x^{2}=-1$$

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

$$2x^{3}+x^{2}+x-5=-1$$

Add 1 to both sides.

$$2x^{3}+x^{2}+x-5+1=0$$

Add -5 and 1 to get -4.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±2,±4,±1,±\frac{1}{2}$$

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 2x^{3}+x^{2}+x-4 by x-1 to get 2x^{2}+3x+4. Solve the equation where the result equals to 0.

$$2x^{2}+3x+4=0$$

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 2 for a, 3 for b, and 4 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\times 4}}{2\times 2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-23}}{4}$$

Solve the equation 2x^{2}+3x+4=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{23}i-3}{4} x=\frac{-3+\sqrt{23}i}{4}$$

Remove the values that the variable cannot be equal to.

$$x\in \emptyset$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{23}i-3}{4} x=\frac{-3+\sqrt{23}i}{4}$$

Variable x cannot be equal to 1.

$$x=\frac{-3+\sqrt{23}i}{4} x=\frac{-\sqrt{23}i-3}{4}$$