Solve 4/x^2-25-x+5/x-5=x-5/x+5-3

$$\frac { 4 } { x ^ { 2 } - 25 } - \frac { x + 5 } { x - 5 } = \frac { x - 5 } { x + 5 } - 3$$

Solve for x

$$x=-11$$

$$x=11$$

Steps Using Difference of Squares

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x^{2}-25,x-5,x+5.

$$4-\left(x+5\right)\left(x+5\right)=\left(x-5\right)\left(x-5\right)+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Multiply x+5 and x+5 to get \left(x+5\right)^{2}.

$$4-\left(x+5\right)^{2}=\left(x-5\right)\left(x-5\right)+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Multiply x-5 and x-5 to get \left(x-5\right)^{2}.

$$4-\left(x+5\right)^{2}=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$4-\left(x^{2}+10x+25\right)=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$4-x^{2}-10x-25=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Subtract 25 from 4 to get -21.

$$-21-x^{2}-10x=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$-21-x^{2}-10x=x^{2}-10x+25+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use the distributive property to multiply x-5 by x+5 and combine like terms.

$$-21-x^{2}-10x=x^{2}-10x+25+\left(x^{2}-25\right)\left(-3\right)$$

Use the distributive property to multiply x^{2}-25 by -3.

$$-21-x^{2}-10x=x^{2}-10x+25-3x^{2}+75$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-21-x^{2}-10x=-2x^{2}-10x+25+75$$

Add 25 and 75 to get 100.

$$-21-x^{2}-10x=-2x^{2}-10x+100$$

Add 2x^{2} to both sides.

$$-21-x^{2}-10x+2x^{2}=-10x+100$$

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

$$-21+x^{2}-10x=-10x+100$$

Add 10x to both sides.

$$-21+x^{2}-10x+10x=100$$

Combine -10x and 10x to get 0.

$$-21+x^{2}=100$$

Subtract 100 from both sides.

$$-21+x^{2}-100=0$$

Subtract 100 from -21 to get -121.

$$-121+x^{2}=0$$

Consider -121+x^{2}. Rewrite -121+x^{2} as x^{2}-11^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-11\right)\left(x+11\right)=0$$

To find equation solutions, solve x-11=0 and x+11=0.

$$x=11 x=-11$$

Steps by Finding Square Root

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x^{2}-25,x-5,x+5.

$$4-\left(x+5\right)\left(x+5\right)=\left(x-5\right)\left(x-5\right)+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Multiply x+5 and x+5 to get \left(x+5\right)^{2}.

$$4-\left(x+5\right)^{2}=\left(x-5\right)\left(x-5\right)+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Multiply x-5 and x-5 to get \left(x-5\right)^{2}.

$$4-\left(x+5\right)^{2}=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$4-\left(x^{2}+10x+25\right)=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$4-x^{2}-10x-25=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Subtract 25 from 4 to get -21.

$$-21-x^{2}-10x=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$-21-x^{2}-10x=x^{2}-10x+25+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use the distributive property to multiply x-5 by x+5 and combine like terms.

$$-21-x^{2}-10x=x^{2}-10x+25+\left(x^{2}-25\right)\left(-3\right)$$

Use the distributive property to multiply x^{2}-25 by -3.

$$-21-x^{2}-10x=x^{2}-10x+25-3x^{2}+75$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-21-x^{2}-10x=-2x^{2}-10x+25+75$$

Add 25 and 75 to get 100.

$$-21-x^{2}-10x=-2x^{2}-10x+100$$

Add 2x^{2} to both sides.

$$-21-x^{2}-10x+2x^{2}=-10x+100$$

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

$$-21+x^{2}-10x=-10x+100$$

Add 10x to both sides.

$$-21+x^{2}-10x+10x=100$$

Combine -10x and 10x to get 0.

$$-21+x^{2}=100$$

Add 21 to both sides.

$$x^{2}=100+21$$

Add 100 and 21 to get 121.

$$x^{2}=121$$

Take the square root of both sides of the equation.

$$x=11 x=-11$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x^{2}-25,x-5,x+5.

$$4-\left(x+5\right)\left(x+5\right)=\left(x-5\right)\left(x-5\right)+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Multiply x+5 and x+5 to get \left(x+5\right)^{2}.

$$4-\left(x+5\right)^{2}=\left(x-5\right)\left(x-5\right)+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Multiply x-5 and x-5 to get \left(x-5\right)^{2}.

$$4-\left(x+5\right)^{2}=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$4-\left(x^{2}+10x+25\right)=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$4-x^{2}-10x-25=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Subtract 25 from 4 to get -21.

$$-21-x^{2}-10x=\left(x-5\right)^{2}+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$-21-x^{2}-10x=x^{2}-10x+25+\left(x-5\right)\left(x+5\right)\left(-3\right)$$

Use the distributive property to multiply x-5 by x+5 and combine like terms.

$$-21-x^{2}-10x=x^{2}-10x+25+\left(x^{2}-25\right)\left(-3\right)$$

Use the distributive property to multiply x^{2}-25 by -3.

$$-21-x^{2}-10x=x^{2}-10x+25-3x^{2}+75$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-21-x^{2}-10x=-2x^{2}-10x+25+75$$

Add 25 and 75 to get 100.

$$-21-x^{2}-10x=-2x^{2}-10x+100$$

Add 2x^{2} to both sides.

$$-21-x^{2}-10x+2x^{2}=-10x+100$$

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

$$-21+x^{2}-10x=-10x+100$$

Add 10x to both sides.

$$-21+x^{2}-10x+10x=100$$

Combine -10x and 10x to get 0.

$$-21+x^{2}=100$$

Subtract 100 from both sides.

$$-21+x^{2}-100=0$$

Subtract 100 from -21 to get -121.

$$-121+x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$x^{2}-121=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -121 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-121\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-121\right)}}{2}$$

Multiply -4 times -121.

$$x=\frac{0±\sqrt{484}}{2}$$

Take the square root of 484.

$$x=\frac{0±22}{2}$$

Now solve the equation x=\frac{0±22}{2} when ± is plus. Divide 22 by 2.

$$x=11$$

Now solve the equation x=\frac{0±22}{2} when ± is minus. Divide -22 by 2.

$$x=-11$$

The equation is now solved.

$$x=11 x=-11$$