https://file.aiquickdraw.com/mathbot/file/1726205387495szwr90qy.jpg

Let's solve the given functions one by one and find their real roots.

### a) \( f(x) = x^8 + 6x^4 - 7 \)

Let \( u = x^4 \), so the equation becomes:

\[
u^2 + 6u - 7 = 0
\]

This is a quadratic equation in \( u \). Using the quadratic formula:

\[
u = \frac{-6 \pm \sqrt{6^2 - 4(1)(-7)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 28}}{2} = \frac{-6 \pm \sqrt{64}}{2} = \frac{-6 \pm 8}{2}
\]

So, the two solutions are:

\[
u = \frac{-6 + 8}{2} = 1 \quad \text{and} \quad u = \frac{-6 - 8}{2} = -7
\]

Now, we substitute back \( u = x^4 \):

1. For \( x^4 = 1 \), we get \( x = \pm 1 \).
2. For \( x^4 = -7 \), there are no real solutions, since \( x^4 \) cannot be negative for real numbers.

Thus, the real roots for part a) are:

\[
x = \pm 1
\]

### b) \( f(x) = 6x^{10} - 5x^5 + 1 \)

Let \( u = x^5 \), so the equation becomes:

\[
6u^2 - 5u + 1 = 0
\]

Using the quadratic formula again:

\[
u = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(6)(1)}}{2(6)} = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm \sqrt{1}}{12} = \frac{5 \pm 1}{12}
\]

The two solutions are:

\[
u = \frac{5 + 1}{12} = \frac{6}{12} = \frac{1}{2} \quad \text{and} \quad u = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3}
\]

Now substitute back \( u = x^5 \):

1. \( x^5 = \frac{1}{2} \) has the real solution \( x = \sqrt[5]{\frac{1}{2}} \).
2. \( x^5 = \frac{1}{3} \) has the real solution \( x = \sqrt[5]{\frac{1}{3}} \).

Thus, the real roots for part b) are:

\[
x = \sqrt[5]{\frac{1}{2}} \quad \text{and} \quad x = \sqrt[5]{\frac{1}{3}}
\]

### c) \( f(x) = x^4 - 3x^{\frac{3}{2}} + 2 \)

To solve this, we set \( y = x^{\frac{1}{2}} \), so \( x = y^2 \) and the equation becomes:

\[
y^8 - 3y^3 + 2 = 0
\]

This is a complex higher-degree polynomial. Finding the real roots may involve numerical methods such as Newton's method, or graphing.

However, by trial or estimation, we can approximate the roots of this equation or factor it if possible. Since it is not easily factored or solved algebraically, we could explore approximate solutions, but exact methods will require advanced techniques or numerical tools.

Would you like to go further with part c) using numerical methods?

Let me know if you want more details on any step!

---

Here are five related questions:

1. How do you solve higher-degree polynomials using substitution?
2. What is the quadratic formula, and how is it derived?
3. What are the properties of polynomials with respect to real roots?
4. How do you determine if a polynomial has no real solutions?
5. What are some methods for solving polynomials that cannot be factored easily?

**Tip**: Always try simplifying equations with substitutions like \( u = x^k \) when dealing with high-degree polynomials.

Find all real roots of the following functions:

a) f(x) = x^8 + 6x^4 - 7
b) f(x) = 6x^10 - 5x^5 + 1
c) f(x) = x^4 - 3x^(3/2) + 2

Solve polynomial equations like f(x) = x^8 + 6x^4 - 7, f(x) = 6x^10 - 5x^5 + 1, and f(x) = x^4 - 3x^(3/2) + 2 by applying algebraic methods, quadratic formulas, and substitution techniques. Ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation