Solve 3/y^2+8/y+2=0

$$\frac { 3 } { y ^ { 2 } } + \frac { 8 } { y } + 2 = 0$$

Solve for y

$$y=\frac{\sqrt{10}}{2}-2\approx -0.41886117$$

$$y=-\frac{\sqrt{10}}{2}-2\approx -3.58113883$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y^{2}, the least common multiple of y^{2},y.

$$3+y\times 8+y^{2}\times 2=0$$

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.

$$2y^{2}+8y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-8±\sqrt{8^{2}-4\times 2\times 3}}{2\times 2}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-8±\sqrt{64-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$y=\frac{-8±\sqrt{64-24}}{2\times 2}$$

Add 64 to -24.

$$y=\frac{-8±\sqrt{40}}{2\times 2}$$

Take the square root of 40.

$$y=\frac{-8±2\sqrt{10}}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-8±2\sqrt{10}}{4}$$

Now solve the equation y=\frac{-8±2\sqrt{10}}{4} when ± is plus. Add -8 to 2\sqrt{10}.

$$y=\frac{2\sqrt{10}-8}{4}$$

Divide -8+2\sqrt{10} by 4.

$$y=\frac{\sqrt{10}}{2}-2$$

Now solve the equation y=\frac{-8±2\sqrt{10}}{4} when ± is minus. Subtract 2\sqrt{10} from -8.

$$y=\frac{-2\sqrt{10}-8}{4}$$

Divide -8-2\sqrt{10} by 4.

$$y=-\frac{\sqrt{10}}{2}-2$$

The equation is now solved.

$$y=\frac{\sqrt{10}}{2}-2 y=-\frac{\sqrt{10}}{2}-2$$

Steps for Completing the Square

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y^{2}, the least common multiple of y^{2},y.

$$3+y\times 8+y^{2}\times 2=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$y\times 8+y^{2}\times 2=-3$$

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.

$$2y^{2}+8y=-3$$

Divide both sides by 2.

$$\frac{2y^{2}+8y}{2}=-\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\frac{8}{2}y=-\frac{3}{2}$$

Divide 8 by 2.

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

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

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

Square 2.

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

Add -\frac{3}{2} to 4.

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

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

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

Take the square root of both sides of the equation.

$$\sqrt{\left(y+2\right)^{2}}=\sqrt{\frac{5}{2}}$$

Simplify.

$$y+2=\frac{\sqrt{10}}{2} y+2=-\frac{\sqrt{10}}{2}$$

Subtract 2 from both sides of the equation.

$$y=\frac{\sqrt{10}}{2}-2 y=-\frac{\sqrt{10}}{2}-2$$