Solve 5^circ/1+x+50/(1+x)^2=100

$$\frac { 5 ^ { \circ } } { 1 + x } + \frac { 50 } { ( 1 + x ) ^ { 2 } } = 100$$

Solve for x

$$x=\frac{-3\sqrt{89}-39}{40}\approx -1.682548585$$

$$x=\frac{3\sqrt{89}-39}{40}\approx -0.267451415$$

Steps Using the Quadratic Formula

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

$$\left(x+1\right)\times 5+50=100\left(x+1\right)^{2}$$

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

$$5x+5+50=100\left(x+1\right)^{2}$$

Add 5 and 50 to get 55.

$$5x+55=100\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$5x+55=100\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply 100 by x^{2}+2x+1.

$$5x+55=100x^{2}+200x+100$$

Subtract 100x^{2} from both sides.

$$5x+55-100x^{2}=200x+100$$

Subtract 200x from both sides.

$$5x+55-100x^{2}-200x=100$$

Combine 5x and -200x to get -195x.

$$-195x+55-100x^{2}=100$$

Subtract 100 from both sides.

$$-195x+55-100x^{2}-100=0$$

Subtract 100 from 55 to get -45.

$$-195x-45-100x^{2}=0$$

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

$$-100x^{2}-195x-45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -100 for a, -195 for b, and -45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-195\right)±\sqrt{\left(-195\right)^{2}-4\left(-100\right)\left(-45\right)}}{2\left(-100\right)}$$

Square -195.

$$x=\frac{-\left(-195\right)±\sqrt{38025-4\left(-100\right)\left(-45\right)}}{2\left(-100\right)}$$

Multiply -4 times -100.

$$x=\frac{-\left(-195\right)±\sqrt{38025+400\left(-45\right)}}{2\left(-100\right)}$$

Multiply 400 times -45.

$$x=\frac{-\left(-195\right)±\sqrt{38025-18000}}{2\left(-100\right)}$$

Add 38025 to -18000.

$$x=\frac{-\left(-195\right)±\sqrt{20025}}{2\left(-100\right)}$$

Take the square root of 20025.

$$x=\frac{-\left(-195\right)±15\sqrt{89}}{2\left(-100\right)}$$

The opposite of -195 is 195.

$$x=\frac{195±15\sqrt{89}}{2\left(-100\right)}$$

Multiply 2 times -100.

$$x=\frac{195±15\sqrt{89}}{-200}$$

Now solve the equation x=\frac{195±15\sqrt{89}}{-200} when ± is plus. Add 195 to 15\sqrt{89}.

$$x=\frac{15\sqrt{89}+195}{-200}$$

Divide 195+15\sqrt{89} by -200.

$$x=\frac{-3\sqrt{89}-39}{40}$$

Now solve the equation x=\frac{195±15\sqrt{89}}{-200} when ± is minus. Subtract 15\sqrt{89} from 195.

$$x=\frac{195-15\sqrt{89}}{-200}$$

Divide 195-15\sqrt{89} by -200.

$$x=\frac{3\sqrt{89}-39}{40}$$

The equation is now solved.

$$x=\frac{-3\sqrt{89}-39}{40} x=\frac{3\sqrt{89}-39}{40}$$

Steps for Completing the Square

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

$$\left(x+1\right)\times 5+50=100\left(x+1\right)^{2}$$

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

$$5x+5+50=100\left(x+1\right)^{2}$$

Add 5 and 50 to get 55.

$$5x+55=100\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$5x+55=100\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply 100 by x^{2}+2x+1.

$$5x+55=100x^{2}+200x+100$$

Subtract 100x^{2} from both sides.

$$5x+55-100x^{2}=200x+100$$

Subtract 200x from both sides.

$$5x+55-100x^{2}-200x=100$$

Combine 5x and -200x to get -195x.

$$-195x+55-100x^{2}=100$$

Subtract 55 from both sides.

$$-195x-100x^{2}=100-55$$

Subtract 55 from 100 to get 45.

$$-195x-100x^{2}=45$$

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.

$$-100x^{2}-195x=45$$

Divide both sides by -100.

$$\frac{-100x^{2}-195x}{-100}=\frac{45}{-100}$$

Dividing by -100 undoes the multiplication by -100.

$$x^{2}+\left(-\frac{195}{-100}\right)x=\frac{45}{-100}$$

Reduce the fraction \frac{-195}{-100} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{39}{20}x=\frac{45}{-100}$$

Reduce the fraction \frac{45}{-100} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{39}{20}x=-\frac{9}{20}$$

Divide \frac{39}{20}, the coefficient of the x term, by 2 to get \frac{39}{40}. Then add the square of \frac{39}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{39}{20}x+\left(\frac{39}{40}\right)^{2}=-\frac{9}{20}+\left(\frac{39}{40}\right)^{2}$$

Square \frac{39}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{39}{20}x+\frac{1521}{1600}=-\frac{9}{20}+\frac{1521}{1600}$$

Add -\frac{9}{20} to \frac{1521}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{39}{20}x+\frac{1521}{1600}=\frac{801}{1600}$$

Factor x^{2}+\frac{39}{20}x+\frac{1521}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{39}{40}\right)^{2}=\frac{801}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{39}{40}\right)^{2}}=\sqrt{\frac{801}{1600}}$$

Simplify.

$$x+\frac{39}{40}=\frac{3\sqrt{89}}{40} x+\frac{39}{40}=-\frac{3\sqrt{89}}{40}$$

Subtract \frac{39}{40} from both sides of the equation.

$$x=\frac{3\sqrt{89}-39}{40} x=\frac{-3\sqrt{89}-39}{40}$$