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

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

Solve for y

$$y = \frac{4 \sqrt{34} + 2}{5} \approx 5.064761516$$

$$y=\frac{2-4\sqrt{34}}{5}\approx -4.264761516$$

Steps Using the Quadratic Formula

Variable y 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(y-5\right)\left(y+5\right), the least common multiple of y^{2}-25,y+5.

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

Use the distributive property to multiply y-5 by 4.

$$3+4y-20=5\left(y-5\right)\left(y+5\right)$$

Subtract 20 from 3 to get -17.

$$-17+4y=5\left(y-5\right)\left(y+5\right)$$

Use the distributive property to multiply 5 by y-5.

$$-17+4y=\left(5y-25\right)\left(y+5\right)$$

Use the distributive property to multiply 5y-25 by y+5 and combine like terms.

$$-17+4y=5y^{2}-125$$

Subtract 5y^{2} from both sides.

$$-17+4y-5y^{2}=-125$$

Add 125 to both sides.

$$-17+4y-5y^{2}+125=0$$

Add -17 and 125 to get 108.

$$108+4y-5y^{2}=0$$

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.

$$-5y^{2}+4y+108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 4 for b, and 108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-4±\sqrt{4^{2}-4\left(-5\right)\times 108}}{2\left(-5\right)}$$

Square 4.

$$y=\frac{-4±\sqrt{16-4\left(-5\right)\times 108}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$y=\frac{-4±\sqrt{16+20\times 108}}{2\left(-5\right)}$$

Multiply 20 times 108.

$$y=\frac{-4±\sqrt{16+2160}}{2\left(-5\right)}$$

Add 16 to 2160.

$$y=\frac{-4±\sqrt{2176}}{2\left(-5\right)}$$

Take the square root of 2176.

$$y=\frac{-4±8\sqrt{34}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$y=\frac{-4±8\sqrt{34}}{-10}$$

Now solve the equation y=\frac{-4±8\sqrt{34}}{-10} when ± is plus. Add -4 to 8\sqrt{34}.

$$y=\frac{8\sqrt{34}-4}{-10}$$

Divide -4+8\sqrt{34} by -10.

$$y=\frac{2-4\sqrt{34}}{5}$$

Now solve the equation y=\frac{-4±8\sqrt{34}}{-10} when ± is minus. Subtract 8\sqrt{34} from -4.

$$y=\frac{-8\sqrt{34}-4}{-10}$$

Divide -4-8\sqrt{34} by -10.

$$y=\frac{4\sqrt{34}+2}{5}$$

The equation is now solved.

$$y=\frac{2-4\sqrt{34}}{5} y=\frac{4\sqrt{34}+2}{5}$$

Steps for Completing the Square

Variable y 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(y-5\right)\left(y+5\right), the least common multiple of y^{2}-25,y+5.

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

Use the distributive property to multiply y-5 by 4.

$$3+4y-20=5\left(y-5\right)\left(y+5\right)$$

Subtract 20 from 3 to get -17.

$$-17+4y=5\left(y-5\right)\left(y+5\right)$$

Use the distributive property to multiply 5 by y-5.

$$-17+4y=\left(5y-25\right)\left(y+5\right)$$

Use the distributive property to multiply 5y-25 by y+5 and combine like terms.

$$-17+4y=5y^{2}-125$$

Subtract 5y^{2} from both sides.

$$-17+4y-5y^{2}=-125$$

Add 17 to both sides.

$$4y-5y^{2}=-125+17$$

Add -125 and 17 to get -108.

$$4y-5y^{2}=-108$$

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.

$$-5y^{2}+4y=-108$$

Divide both sides by -5.

$$\frac{-5y^{2}+4y}{-5}=-\frac{108}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$y^{2}+\frac{4}{-5}y=-\frac{108}{-5}$$

Divide 4 by -5.

$$y^{2}-\frac{4}{5}y=-\frac{108}{-5}$$

Divide -108 by -5.

$$y^{2}-\frac{4}{5}y=\frac{108}{5}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{4}{5}y+\left(-\frac{2}{5}\right)^{2}=\frac{108}{5}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{4}{5}y+\frac{4}{25}=\frac{108}{5}+\frac{4}{25}$$

Add \frac{108}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{4}{5}y+\frac{4}{25}=\frac{544}{25}$$

Factor y^{2}-\frac{4}{5}y+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{2}{5}\right)^{2}=\frac{544}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{2}{5}\right)^{2}}=\sqrt{\frac{544}{25}}$$

Simplify.

$$y-\frac{2}{5}=\frac{4\sqrt{34}}{5} y-\frac{2}{5}=-\frac{4\sqrt{34}}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$y=\frac{4\sqrt{34}+2}{5} y=\frac{2-4\sqrt{34}}{5}$$