To solve the equation \( y^4 - 15y^2 = 16 \) using the quadratic form, we can proceed as follows:

### Step 1: Substitution
Notice that \( y^4 \) can be written as \( (y^2)^2 \). Let's make a substitution to simplify the equation:

Let \( u = y^2 \).

This transforms the equation into a quadratic form:

\[
u^2 - 15u = 16
\]

### Step 2: Rewrite the Equation
Now, rewrite the equation in standard quadratic form:

\[
u^2 - 15u - 16 = 0
\]

### Step 3: Solve the Quadratic Equation
We can solve the quadratic equation using the quadratic formula:

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -15 \), and \( c = -16 \). Plugging in these values:

\[
u = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(-16)}}{2(1)}
\]

\[
u = \frac{15 \pm \sqrt{225 + 64}}{2}
\]

\[
u = \frac{15 \pm \sqrt{289}}{2}
\]

\[
u = \frac{15 \pm 17}{2}
\]

This gives us two possible values for \( u \):

\[
u_1 = \frac{15 + 17}{2} = \frac{32}{2} = 16
\]
\[
u_2 = \frac{15 - 17}{2} = \frac{-2}{2} = -1
\]

### Step 4: Back-Substitution
Recall that \( u = y^2 \). So, we have two equations to solve:

1. \( y^2 = 16 \)
2. \( y^2 = -1 \)

### Step 5: Solve for \( y \)
For the first equation:

\[
y^2 = 16 \implies y = \pm \sqrt{16} = \pm 4
\]

For the second equation:

\[
y^2 = -1 \implies y = \pm \sqrt{-1} = \pm i
\]

### Final Answer:
The solutions for \( y \) are:

\[
y = \pm 4, \quad y = \pm i
\]

So, the complete set of solutions is:

\[
y = 4, \, -4, \, i, \, -i
\]

Would you like more details on any part of this solution? 

Here are 5 related questions:
1. How do you derive the quadratic formula?
2. What are the applications of complex numbers?
3. How do you solve quadratic equations by factoring?
4. What are other methods to solve quadratic equations?
5. How does the discriminant determine the nature of solutions?

**Tip:** When solving polynomial equations, substitution can often simplify the problem, especially when dealing with higher powers.

Learn how to solve the equation y⁴ - 15y² = 16 step-by-step using the quadratic form. This problem involves quadratic equations, complex numbers, and the fundamental theorem of algebra. Suitable for students in Grades 11-12.

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