Solve (y^2/4)^2+y^2=5

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

Solve for y

$$y=-2$$

$$y=2$$

Steps Using the Quadratic Formula

To raise \frac{y^{2}}{4} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply y^{2} times \frac{4^{2}}{4^{2}}.

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

Since \frac{\left(y^{2}\right)^{2}}{4^{2}} and \frac{y^{2}\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Calculate 4 to the power of 2 and get 16.

$$\frac{y^{4}+y^{2}\times 16}{4^{2}}=5$$

Calculate 4 to the power of 2 and get 16.

$$\frac{y^{4}+y^{2}\times 16}{16}=5$$

Divide each term of y^{4}+y^{2}\times 16 by 16 to get \frac{1}{16}y^{4}+y^{2}.

$$\frac{1}{16}y^{4}+y^{2}=5$$

Subtract 5 from both sides.

$$\frac{1}{16}y^{4}+y^{2}-5=0$$

Substitute t for y^{2}.

$$\frac{1}{16}t^{2}+t-5=0$$

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 \frac{1}{16} for a, 1 for b, and -5 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{16}\left(-5\right)}}{\frac{1}{16}\times 2}$$

Do the calculations.

$$t=\frac{-1±\frac{3}{2}}{\frac{1}{8}}$$

Solve the equation t=\frac{-1±\frac{3}{2}}{\frac{1}{8}} when ± is plus and when ± is minus.

$$t=4 t=-20$$

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

$$y=2 y=-2$$