proof that if the quadratic function has no zeros then the truth is that a(2a + 3b + 6c) > 0

Let's first break down the problem and the proof logically.

We are given a quadratic function in the standard form:

\[
f(x) = ax^2 + bx + c
\]

We need to prove that if the quadratic function has **no zeros**, the inequality \( a(2a + 3b + 6c) > 0 \) holds true.

### Step 1: Condition for No Zeros

A quadratic function \( f(x) = ax^2 + bx + c \) has no real zeros if and only if its **discriminant** is negative. The discriminant \( \Delta \) of the quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[
\Delta = b^2 - 4ac
\]

If \( \Delta < 0 \), the quadratic function has no real roots.

### Step 2: Analyzing \( a(2a + 3b + 6c) \)

We are tasked with proving that if \( \Delta < 0 \), then the inequality \( a(2a + 3b + 6c) > 0 \) holds.

Let's express and investigate \( a(2a + 3b + 6c) \):

\[
a(2a + 3b + 6c) = 2a^2 + 3ab + 6ac
\]

Thus, we need to show that if \( \Delta = b^2 - 4ac < 0 \), then:

\[
2a^2 + 3ab + 6ac > 0
\]

### Step 3: Considering \( a > 0 \) and \( a < 0 \)

#### Case 1: \( a > 0 \)

When \( a > 0 \), the parabola opens upwards. If there are no real roots (i.e., \( \Delta < 0 \)), the quadratic stays above the \( x \)-axis, meaning that \( f(x) > 0 \) for all \( x \). In this case, \( f(x) \) is always positive, and we expect \( a(2a + 3b + 6c) > 0 \).

#### Case 2: \( a < 0 \)

When \( a < 0 \), the parabola opens downwards. If there are no real roots (i.e., \( \Delta < 0 \)), the quadratic stays below the \( x \)-axis, meaning that \( f(x) < 0 \) for all \( x \). In this case, we expect that \( a(2a + 3b + 6c) > 0 \) because the product of negative terms would still yield a positive result.

### Step 4: Conclusion

Since \( \Delta < 0 \) implies no real roots, and depending on the sign of \( a \), either \( f(x) > 0 \) (if \( a > 0 \)) or \( f(x) < 0 \) (if \( a < 0 \)), we conclude that:

\[
a(2a + 3b + 6c) > 0
\]

Hence, the proof holds true.

### Summary:

The quadratic having no real zeros implies \( a(2a + 3b + 6c) > 0 \) due to the behavior of the function depending on the sign of \( a \). In both cases (whether \( a > 0 \) or \( a < 0 \)), the inequality is satisfied.

Would you like more details on any step?

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Here are 5 related questions for deeper understanding:
1. How does the discriminant determine the number of real roots for a quadratic function?
2. What happens graphically when a quadratic has no real roots?
3. Can you derive the expression for the vertex of a quadratic function?
4. What is the role of the coefficients in determining the shape of the quadratic curve?
5. How can you analyze the inequality using the vertex form of the quadratic?

**Tip**: The discriminant is key in understanding the nature of roots in any quadratic equation. Always check it first when analyzing the behavior of the quadratic!

This guide explores the proof that for a quadratic function with no real zeros, the inequality a(2a + 3b + 6c) > 0 holds true. Learn how the discriminant and quadratic function properties come into play in this advanced algebraic problem. Suitable for students in Grades 10-12.

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