Solve 1/4x^2-20x+25+121/4x^2+20x+25=22/4x^2-25

$$\frac{ 1 }{ 4 { x }^{ 2 } -20x+25 } + \frac{ 121 }{ 4 { x }^{ 2 } +20x+25 } = \frac{ 22 }{ 4 { x }^{ 2 } -25 }$$

Solve for x

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{5}{2},\frac{5}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-5\right)^{2}\left(2x+5\right)^{2}, the least common multiple of 4x^{2}-20x+25,4x^{2}+20x+25,4x^{2}-25.

$$\left(2x+5\right)^{2}+\left(2x-5\right)^{2}\times 121=\left(4x^{2}-25\right)\times 22$$

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

$$4x^{2}+20x+25+\left(2x-5\right)^{2}\times 121=\left(4x^{2}-25\right)\times 22$$

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

$$4x^{2}+20x+25+\left(4x^{2}-20x+25\right)\times 121=\left(4x^{2}-25\right)\times 22$$

Use the distributive property to multiply 4x^{2}-20x+25 by 121.

$$4x^{2}+20x+25+484x^{2}-2420x+3025=\left(4x^{2}-25\right)\times 22$$

Combine 4x^{2} and 484x^{2} to get 488x^{2}.

$$488x^{2}+20x+25-2420x+3025=\left(4x^{2}-25\right)\times 22$$

Combine 20x and -2420x to get -2400x.

$$488x^{2}-2400x+25+3025=\left(4x^{2}-25\right)\times 22$$

Add 25 and 3025 to get 3050.

$$488x^{2}-2400x+3050=\left(4x^{2}-25\right)\times 22$$

Use the distributive property to multiply 4x^{2}-25 by 22.

$$488x^{2}-2400x+3050=88x^{2}-550$$

Subtract 88x^{2} from both sides.

$$488x^{2}-2400x+3050-88x^{2}=-550$$

Combine 488x^{2} and -88x^{2} to get 400x^{2}.

$$400x^{2}-2400x+3050=-550$$

Add 550 to both sides.

$$400x^{2}-2400x+3050+550=0$$

Add 3050 and 550 to get 3600.

$$400x^{2}-2400x+3600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 400 for a, -2400 for b, and 3600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2400\right)±\sqrt{\left(-2400\right)^{2}-4\times 400\times 3600}}{2\times 400}$$

Square -2400.

$$x=\frac{-\left(-2400\right)±\sqrt{5760000-4\times 400\times 3600}}{2\times 400}$$

Multiply -4 times 400.

$$x=\frac{-\left(-2400\right)±\sqrt{5760000-1600\times 3600}}{2\times 400}$$

Multiply -1600 times 3600.

$$x=\frac{-\left(-2400\right)±\sqrt{5760000-5760000}}{2\times 400}$$

Add 5760000 to -5760000.

$$x=\frac{-\left(-2400\right)±\sqrt{0}}{2\times 400}$$

Take the square root of 0.

$$x=-\frac{-2400}{2\times 400}$$

The opposite of -2400 is 2400.

$$x=\frac{2400}{2\times 400}$$

Multiply 2 times 400.

$$x=\frac{2400}{800}$$

Divide 2400 by 800.

$$x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{5}{2},\frac{5}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-5\right)^{2}\left(2x+5\right)^{2}, the least common multiple of 4x^{2}-20x+25,4x^{2}+20x+25,4x^{2}-25.

$$\left(2x+5\right)^{2}+\left(2x-5\right)^{2}\times 121=\left(4x^{2}-25\right)\times 22$$

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

$$4x^{2}+20x+25+\left(2x-5\right)^{2}\times 121=\left(4x^{2}-25\right)\times 22$$

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

$$4x^{2}+20x+25+\left(4x^{2}-20x+25\right)\times 121=\left(4x^{2}-25\right)\times 22$$

Use the distributive property to multiply 4x^{2}-20x+25 by 121.

$$4x^{2}+20x+25+484x^{2}-2420x+3025=\left(4x^{2}-25\right)\times 22$$

Combine 4x^{2} and 484x^{2} to get 488x^{2}.

$$488x^{2}+20x+25-2420x+3025=\left(4x^{2}-25\right)\times 22$$

Combine 20x and -2420x to get -2400x.

$$488x^{2}-2400x+25+3025=\left(4x^{2}-25\right)\times 22$$

Add 25 and 3025 to get 3050.

$$488x^{2}-2400x+3050=\left(4x^{2}-25\right)\times 22$$

Use the distributive property to multiply 4x^{2}-25 by 22.

$$488x^{2}-2400x+3050=88x^{2}-550$$

Subtract 88x^{2} from both sides.

$$488x^{2}-2400x+3050-88x^{2}=-550$$

Combine 488x^{2} and -88x^{2} to get 400x^{2}.

$$400x^{2}-2400x+3050=-550$$

Subtract 3050 from both sides.

$$400x^{2}-2400x=-550-3050$$

Subtract 3050 from -550 to get -3600.

$$400x^{2}-2400x=-3600$$

Divide both sides by 400.

$$\frac{400x^{2}-2400x}{400}=-\frac{3600}{400}$$

Dividing by 400 undoes the multiplication by 400.

$$x^{2}+\left(-\frac{2400}{400}\right)x=-\frac{3600}{400}$$

Divide -2400 by 400.

$$x^{2}-6x=-\frac{3600}{400}$$

Divide -3600 by 400.

$$x^{2}-6x=-9$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -3.

$$x^{2}-6x+9=-9+9$$

Add -9 to 9.

$$x^{2}-6x+9=0$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-3=0 x-3=0$$

Add 3 to both sides of the equation.

$$x=3 x=3$$

The equation is now solved. Solutions are the same.

$$x=3$$