Solve 144/y^2+y^2=40

$$\frac { 144 } { y ^ { 2 } } + y ^ { 2 } = 40$$

Solve for y

$$y=-2$$

$$y=2$$

$$y=6$$

$$y=-6$$

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}.

$$144+y^{2}y^{2}=40y^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$144+y^{4}=40y^{2}$$

Subtract 40y^{2} from both sides.

$$144+y^{4}-40y^{2}=0$$

Substitute t for y^{2}.

$$t^{2}-40t+144=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -40 for b, and 144 for c in the quadratic formula.

$$t=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 1\times 144}}{2}$$

Do the calculations.

$$t=\frac{40±32}{2}$$

Solve the equation t=\frac{40±32}{2} when ± is plus and when ± is minus.

$$t=36 t=4$$

Since y=t^{2}, the solutions are obtained by evaluating y=±\sqrt{t} for each t.

$$y=6 y=-6 y=2 y=-2$$