Solve x^2+1/x-4-x^2-1/x+3=23

$$\frac{ { x }^{ 2 } +1 }{ x-4 } - \frac{ { x }^{ 2 } -1 }{ x+3 } = 23$$

Solve for x

$$x = -\frac{55}{16} = -3\frac{7}{16} = -3.4375$$

$$x=5$$

Steps Using Factoring By Grouping

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

$$\left(x+3\right)\left(x^{2}+1\right)-\left(x-4\right)\left(x^{2}-1\right)=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply x+3 by x^{2}+1.

$$x^{3}+x+3x^{2}+3-\left(x-4\right)\left(x^{2}-1\right)=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply x-4 by x^{2}-1.

$$x^{3}+x+3x^{2}+3-\left(x^{3}-x-4x^{2}+4\right)=23\left(x-4\right)\left(x+3\right)$$

To find the opposite of x^{3}-x-4x^{2}+4, find the opposite of each term.

$$x^{3}+x+3x^{2}+3-x^{3}+x+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine x^{3} and -x^{3} to get 0.

$$x+3x^{2}+3+x+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine x and x to get 2x.

$$2x+3x^{2}+3+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$2x+7x^{2}+3-4=23\left(x-4\right)\left(x+3\right)$$

Subtract 4 from 3 to get -1.

$$2x+7x^{2}-1=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply 23 by x-4.

$$2x+7x^{2}-1=\left(23x-92\right)\left(x+3\right)$$

Use the distributive property to multiply 23x-92 by x+3 and combine like terms.

$$2x+7x^{2}-1=23x^{2}-23x-276$$

Subtract 23x^{2} from both sides.

$$2x+7x^{2}-1-23x^{2}=-23x-276$$

Combine 7x^{2} and -23x^{2} to get -16x^{2}.

$$2x-16x^{2}-1=-23x-276$$

Add 23x to both sides.

$$2x-16x^{2}-1+23x=-276$$

Combine 2x and 23x to get 25x.

$$25x-16x^{2}-1=-276$$

Add 276 to both sides.

$$25x-16x^{2}-1+276=0$$

Add -1 and 276 to get 275.

$$25x-16x^{2}+275=0$$

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

$$-16x^{2}+25x+275=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -16x^{2}+ax+bx+275. To find a and b, set up a system to be solved.

$$a+b=25 ab=-16\times 275=-4400$$

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 -4400.

$$-1,4400 -2,2200 -4,1100 -5,880 -8,550 -10,440 -11,400 -16,275 -20,220 -22,200 -25,176 -40,110 -44,100 -50,88 -55,80$$

Calculate the sum for each pair.

$$-1+4400=4399 -2+2200=2198 -4+1100=1096 -5+880=875 -8+550=542 -10+440=430 -11+400=389 -16+275=259 -20+220=200 -22+200=178 -25+176=151 -40+110=70 -44+100=56 -50+88=38 -55+80=25$$

The solution is the pair that gives sum 25.

$$a=80 b=-55$$

Rewrite -16x^{2}+25x+275 as \left(-16x^{2}+80x\right)+\left(-55x+275\right).

$$\left(-16x^{2}+80x\right)+\left(-55x+275\right)$$

Factor out 16x in the first and 55 in the second group.

$$16x\left(-x+5\right)+55\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(16x+55\right)$$

To find equation solutions, solve -x+5=0 and 16x+55=0.

$$x=5 x=-\frac{55}{16}$$

Steps Using the Quadratic Formula

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

$$\left(x+3\right)\left(x^{2}+1\right)-\left(x-4\right)\left(x^{2}-1\right)=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply x+3 by x^{2}+1.

$$x^{3}+x+3x^{2}+3-\left(x-4\right)\left(x^{2}-1\right)=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply x-4 by x^{2}-1.

$$x^{3}+x+3x^{2}+3-\left(x^{3}-x-4x^{2}+4\right)=23\left(x-4\right)\left(x+3\right)$$

To find the opposite of x^{3}-x-4x^{2}+4, find the opposite of each term.

$$x^{3}+x+3x^{2}+3-x^{3}+x+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine x^{3} and -x^{3} to get 0.

$$x+3x^{2}+3+x+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine x and x to get 2x.

$$2x+3x^{2}+3+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$2x+7x^{2}+3-4=23\left(x-4\right)\left(x+3\right)$$

Subtract 4 from 3 to get -1.

$$2x+7x^{2}-1=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply 23 by x-4.

$$2x+7x^{2}-1=\left(23x-92\right)\left(x+3\right)$$

Use the distributive property to multiply 23x-92 by x+3 and combine like terms.

$$2x+7x^{2}-1=23x^{2}-23x-276$$

Subtract 23x^{2} from both sides.

$$2x+7x^{2}-1-23x^{2}=-23x-276$$

Combine 7x^{2} and -23x^{2} to get -16x^{2}.

$$2x-16x^{2}-1=-23x-276$$

Add 23x to both sides.

$$2x-16x^{2}-1+23x=-276$$

Combine 2x and 23x to get 25x.

$$25x-16x^{2}-1=-276$$

Add 276 to both sides.

$$25x-16x^{2}-1+276=0$$

Add -1 and 276 to get 275.

$$25x-16x^{2}+275=0$$

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.

$$-16x^{2}+25x+275=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 25 for b, and 275 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-16\right)\times 275}}{2\left(-16\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-16\right)\times 275}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-25±\sqrt{625+64\times 275}}{2\left(-16\right)}$$

Multiply 64 times 275.

$$x=\frac{-25±\sqrt{625+17600}}{2\left(-16\right)}$$

Add 625 to 17600.

$$x=\frac{-25±\sqrt{18225}}{2\left(-16\right)}$$

Take the square root of 18225.

$$x=\frac{-25±135}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-25±135}{-32}$$

Now solve the equation x=\frac{-25±135}{-32} when ± is plus. Add -25 to 135.

$$x=\frac{110}{-32}$$

Reduce the fraction \frac{110}{-32} to lowest terms by extracting and canceling out 2.

$$x=-\frac{55}{16}$$

Now solve the equation x=\frac{-25±135}{-32} when ± is minus. Subtract 135 from -25.

$$x=-\frac{160}{-32}$$

Divide -160 by -32.

$$x=5$$

The equation is now solved.

$$x=-\frac{55}{16} x=5$$

Steps for Completing the Square

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

$$\left(x+3\right)\left(x^{2}+1\right)-\left(x-4\right)\left(x^{2}-1\right)=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply x+3 by x^{2}+1.

$$x^{3}+x+3x^{2}+3-\left(x-4\right)\left(x^{2}-1\right)=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply x-4 by x^{2}-1.

$$x^{3}+x+3x^{2}+3-\left(x^{3}-x-4x^{2}+4\right)=23\left(x-4\right)\left(x+3\right)$$

To find the opposite of x^{3}-x-4x^{2}+4, find the opposite of each term.

$$x^{3}+x+3x^{2}+3-x^{3}+x+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine x^{3} and -x^{3} to get 0.

$$x+3x^{2}+3+x+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine x and x to get 2x.

$$2x+3x^{2}+3+4x^{2}-4=23\left(x-4\right)\left(x+3\right)$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$2x+7x^{2}+3-4=23\left(x-4\right)\left(x+3\right)$$

Subtract 4 from 3 to get -1.

$$2x+7x^{2}-1=23\left(x-4\right)\left(x+3\right)$$

Use the distributive property to multiply 23 by x-4.

$$2x+7x^{2}-1=\left(23x-92\right)\left(x+3\right)$$

Use the distributive property to multiply 23x-92 by x+3 and combine like terms.

$$2x+7x^{2}-1=23x^{2}-23x-276$$

Subtract 23x^{2} from both sides.

$$2x+7x^{2}-1-23x^{2}=-23x-276$$

Combine 7x^{2} and -23x^{2} to get -16x^{2}.

$$2x-16x^{2}-1=-23x-276$$

Add 23x to both sides.

$$2x-16x^{2}-1+23x=-276$$

Combine 2x and 23x to get 25x.

$$25x-16x^{2}-1=-276$$

Add 1 to both sides.

$$25x-16x^{2}=-276+1$$

Add -276 and 1 to get -275.

$$25x-16x^{2}=-275$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-16x^{2}+25x=-275$$

Divide both sides by -16.

$$\frac{-16x^{2}+25x}{-16}=-\frac{275}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$x^{2}+\frac{25}{-16}x=-\frac{275}{-16}$$

Divide 25 by -16.

$$x^{2}-\frac{25}{16}x=-\frac{275}{-16}$$

Divide -275 by -16.

$$x^{2}-\frac{25}{16}x=\frac{275}{16}$$

Divide -\frac{25}{16}, the coefficient of the x term, by 2 to get -\frac{25}{32}. Then add the square of -\frac{25}{32} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{16}x+\left(-\frac{25}{32}\right)^{2}=\frac{275}{16}+\left(-\frac{25}{32}\right)^{2}$$

Square -\frac{25}{32} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{16}x+\frac{625}{1024}=\frac{275}{16}+\frac{625}{1024}$$

Add \frac{275}{16} to \frac{625}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{25}{16}x+\frac{625}{1024}=\frac{18225}{1024}$$

Factor x^{2}-\frac{25}{16}x+\frac{625}{1024}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{32}\right)^{2}=\frac{18225}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{32}\right)^{2}}=\sqrt{\frac{18225}{1024}}$$

Simplify.

$$x-\frac{25}{32}=\frac{135}{32} x-\frac{25}{32}=-\frac{135}{32}$$

Add \frac{25}{32} to both sides of the equation.

$$x=5 x=-\frac{55}{16}$$