Solve 2x^2/x+1-5/x-3+x^2+11/x^2-2x-3=0

$$\frac { 2 x ^ { 2 } } { x + 1 } - \frac { 5 } { x - 3 } + \frac { x ^ { 2 } + 11 } { x ^ { 2 } - 2 x - 3 } = 0$$

Solve for x

$$x=\frac{\sqrt{17}-1}{4}\approx 0.780776406$$

$$x=\frac{-\sqrt{17}-1}{4}\approx -1.280776406$$

Solution Steps

Variable x cannot be equal to any of the values -1,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+1\right), the least common multiple of x+1,x-3,x^{2}-2x-3.

$$\left(x-3\right)\times 2x^{2}-\left(x+1\right)\times 5+x^{2}+11=0$$

Use the distributive property to multiply x-3 by 2.

$$\left(2x-6\right)x^{2}-\left(x+1\right)\times 5+x^{2}+11=0$$

Use the distributive property to multiply 2x-6 by x^{2}.

$$2x^{3}-6x^{2}-\left(x+1\right)\times 5+x^{2}+11=0$$

Use the distributive property to multiply x+1 by 5.

$$2x^{3}-6x^{2}-\left(5x+5\right)+x^{2}+11=0$$

To find the opposite of 5x+5, find the opposite of each term.

$$2x^{3}-6x^{2}-5x-5+x^{2}+11=0$$

Combine -6x^{2} and x^{2} to get -5x^{2}.

$$2x^{3}-5x^{2}-5x-5+11=0$$

Add -5 and 11 to get 6.

$$2x^{3}-5x^{2}-5x+6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-5x^{2}-5x+6 by x-3 to get 2x^{2}+x-2. Solve the equation where the result equals to 0.

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

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, 1 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±\sqrt{17}}{4}$$

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

$$x=\frac{-\sqrt{17}-1}{4} x=\frac{\sqrt{17}-1}{4}$$

Remove the values that the variable cannot be equal to.

$$x\in \emptyset$$

List all found solutions.

$$x=3 x=\frac{-\sqrt{17}-1}{4} x=\frac{\sqrt{17}-1}{4}$$

Variable x cannot be equal to 3.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$