Let's solve the given equation step by step:

### Equation:
\[
25x^4 - 56x^2 + 49(4) = 0
\]

### Step 1: Simplify the equation
Simplify the constant term \( 49(4) \).

\[
49 \times 4 = 196
\]

So, the equation becomes:
\[
25x^4 - 56x^2 + 196 = 0
\]

### Step 2: Introduce substitution
Let \( y = x^2 \), which means \( x^4 = y^2 \). Now, rewrite the equation in terms of \( y \):

\[
25y^2 - 56y + 196 = 0
\]

### Step 3: Solve the quadratic equation
This is a quadratic equation in terms of \( y \). We will use the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( 25y^2 - 56y + 196 = 0 \), we have:
- \( a = 25 \)
- \( b = -56 \)
- \( c = 196 \)

Now, apply the values to the quadratic formula:

\[
y = \frac{-(-56) \pm \sqrt{(-56)^2 - 4(25)(196)}}{2(25)}
\]

Simplify:

\[
y = \frac{56 \pm \sqrt{3136 - 19600}}{50}
\]

\[
y = \frac{56 \pm \sqrt{-16464}}{50}
\]

Since the discriminant is negative, the equation has no real solutions. Therefore, the quadratic equation has **no real roots**.

### Conclusion:
The equation \( 25x^4 - 56x^2 + 196 = 0 \) has no real solutions because the discriminant is negative.

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. What is a discriminant and how does it determine the nature of the roots?
2. How do you solve quadratic equations when the discriminant is negative?
3. Can the original equation have complex solutions?
4. How does substitution simplify higher-degree polynomial equations?
5. How would the solution change if the constant term were different?

**Tip:** For equations involving squares and higher powers, substitution can simplify the solving process by reducing the degree of the equation.

Discover how to solve the polynomial equation 25x^4 - 56x^2 + 49(4) using substitution and the quadratic formula. Explore key algebraic concepts like the discriminant and learn why this equation has no real solutions. Ideal for students in Grades 10-12.

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