Solve 4/x^2-1+15+7x/x^4-5x^2+4=2/4-x^2

$$\frac { 4 } { x ^ { 2 } - 1 } + \frac { 15 + 7 x } { x ^ { 4 } - 5 x ^ { 2 } + 4 } = \frac { 2 } { 4 - x ^ { 2 } }$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,-1,1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right), the least common multiple of x^{2}-1,x^{4}-5x^{2}+4,4-x^{2}.

$$\left(x^{2}-4\right)\times 4+15+7x=\left(-x^{2}+1\right)\times 2$$

Use the distributive property to multiply x^{2}-4 by 4.

$$4x^{2}-16+15+7x=\left(-x^{2}+1\right)\times 2$$

Add -16 and 15 to get -1.

$$4x^{2}-1+7x=\left(-x^{2}+1\right)\times 2$$

Use the distributive property to multiply -x^{2}+1 by 2.

$$4x^{2}-1+7x=-2x^{2}+2$$

Add 2x^{2} to both sides.

$$4x^{2}-1+7x+2x^{2}=2$$

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$6x^{2}-1+7x=2$$

Subtract 2 from both sides.

$$6x^{2}-1+7x-2=0$$

Subtract 2 from -1 to get -3.

$$6x^{2}-3+7x=0$$

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

$$6x^{2}+7x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=7 ab=6\left(-3\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 7.

$$a=-2 b=9$$

Rewrite 6x^{2}+7x-3 as \left(6x^{2}-2x\right)+\left(9x-3\right).

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

Factor out 2x in the first and 3 in the second group.

$$2x\left(3x-1\right)+3\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(2x+3\right)$$

To find equation solutions, solve 3x-1=0 and 2x+3=0.

$$x=\frac{1}{3} x=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,-1,1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right), the least common multiple of x^{2}-1,x^{4}-5x^{2}+4,4-x^{2}.

$$\left(x^{2}-4\right)\times 4+15+7x=\left(-x^{2}+1\right)\times 2$$

Use the distributive property to multiply x^{2}-4 by 4.

$$4x^{2}-16+15+7x=\left(-x^{2}+1\right)\times 2$$

Add -16 and 15 to get -1.

$$4x^{2}-1+7x=\left(-x^{2}+1\right)\times 2$$

Use the distributive property to multiply -x^{2}+1 by 2.

$$4x^{2}-1+7x=-2x^{2}+2$$

Add 2x^{2} to both sides.

$$4x^{2}-1+7x+2x^{2}=2$$

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$6x^{2}-1+7x=2$$

Subtract 2 from both sides.

$$6x^{2}-1+7x-2=0$$

Subtract 2 from -1 to get -3.

$$6x^{2}-3+7x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$6x^{2}+7x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 6\left(-3\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-7±\sqrt{49-24\left(-3\right)}}{2\times 6}$$

Multiply -24 times -3.

$$x=\frac{-7±\sqrt{49+72}}{2\times 6}$$

Add 49 to 72.

$$x=\frac{-7±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-7±11}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-7±11}{12}$$

Now solve the equation x=\frac{-7±11}{12} when ± is plus. Add -7 to 11.

$$x=\frac{4}{12}$$

Reduce the fraction \frac{4}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-7±11}{12} when ± is minus. Subtract 11 from -7.

$$x=-\frac{18}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=\frac{1}{3} x=-\frac{3}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,-1,1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right), the least common multiple of x^{2}-1,x^{4}-5x^{2}+4,4-x^{2}.

$$\left(x^{2}-4\right)\times 4+15+7x=\left(-x^{2}+1\right)\times 2$$

Use the distributive property to multiply x^{2}-4 by 4.

$$4x^{2}-16+15+7x=\left(-x^{2}+1\right)\times 2$$

Add -16 and 15 to get -1.

$$4x^{2}-1+7x=\left(-x^{2}+1\right)\times 2$$

Use the distributive property to multiply -x^{2}+1 by 2.

$$4x^{2}-1+7x=-2x^{2}+2$$

Add 2x^{2} to both sides.

$$4x^{2}-1+7x+2x^{2}=2$$

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$6x^{2}-1+7x=2$$

Add 1 to both sides.

$$6x^{2}+7x=2+1$$

Add 2 and 1 to get 3.

$$6x^{2}+7x=3$$

Divide both sides by 6.

$$\frac{6x^{2}+7x}{6}=\frac{3}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{7}{6}x=\frac{3}{6}$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{7}{6}x=\frac{1}{2}$$

Divide \frac{7}{6}, the coefficient of the x term, by 2 to get \frac{7}{12}. Then add the square of \frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{6}x+\left(\frac{7}{12}\right)^{2}=\frac{1}{2}+\left(\frac{7}{12}\right)^{2}$$

Square \frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{6}x+\frac{49}{144}=\frac{1}{2}+\frac{49}{144}$$

Add \frac{1}{2} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{6}x+\frac{49}{144}=\frac{121}{144}$$

Factor x^{2}+\frac{7}{6}x+\frac{49}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{12}\right)^{2}=\frac{121}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{12}\right)^{2}}=\sqrt{\frac{121}{144}}$$

Simplify.

$$x+\frac{7}{12}=\frac{11}{12} x+\frac{7}{12}=-\frac{11}{12}$$

Subtract \frac{7}{12} from both sides of the equation.

$$x=\frac{1}{3} x=-\frac{3}{2}$$