Problem 29 - Algebra 2

Question

Find all of the zeros of each function. \(h(x)=4 x^{4}+17 x^{2}+4\)

Step-by-Step Solution

Answer

The function has no real zeros.

Step 1: Set the Function to Zero

The problem asks us to find the zeros of the function, which are the values of \(x\) that make the function equal to zero. Begin by setting \(h(x)\) equal to zero: \[4x^4 + 17x^2 + 4 = 0\]

Step 2: Substitute Variables

Notice that the equation resembles a quadratic form, where \(4x^4\) can be thought of as \((2x^2)^2\). We use substitution by letting \(y = x^2\). Thus, the equation becomes \[4y^2 + 17y + 4 = 0\].

Step 3: Solve the Quadratic Equation

Now, solve the quadratic equation for \(y\) using the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a = 4\), \(b = 17\), and \(c = 4\). Thus, \[y = \frac{-17 \pm \sqrt{17^2 - 4(4)(4)}}{2(4)}\]. Simplify the terms inside the square root: \[y = \frac{-17 \pm \sqrt{289 - 64}}{8}\] \[y = \frac{-17 \pm \sqrt{225}}{8}\].

Step 4: Simplify the Solutions for y

Continue simplifying: \[y = \frac{-17 \pm 15}{8}\]. This gives two solutions: \[y_1 = \frac{-17 + 15}{8} = \frac{-2}{8} = -\frac{1}{4}\] and \[y_2 = \frac{-17 - 15}{8} = \frac{-32}{8} = -4\].

Step 5: Back-Substitute to Find x

Since \(y = x^2\), back-substitute to find \(x\). For \(y_1 = -\frac{1}{4}\), we have \(x^2 = -\frac{1}{4}\), which has no real solutions as a square of a real number cannot be negative. For \(y_2 = -4\), \(x^2 = -4\) also has no real solutions for the same reason.

Step 6: Conclusion

Since both values of \(y\) lead to imaginary solutions for \(x\), \(h(x) = 4x^4 + 17x^2 + 4\) has no real zeros.

Key Concepts

Zeros of a Function

The zeros of a function are essentially the values of the variable that make the function equal to zero. For a given function, finding its zeros is akin to solving an equation where the function is set to zero. Think of zeros as the points where the graph of the function touches the x-axis. These are important in determining the roots, solutions, or intercepts of the function.
To find these zeros, you initially set the function equal to zero. This turns our original problem into a search for the particular values of the variable that satisfy this condition.

• For example, if our function is expressed as \(f(x) = ax^2 + bx + c\), we would set \(f(x) = 0\) to find its zeros.
• Solving this yields the points where the function crosses or touches the x-axis.

It is crucial to understand that not all functions have real zeros. As seen in the exercise with the function \(h(x) = 4x^4 + 17x^2 + 4\), after analysis, it was determined to have no real zeros.

Substitution Method

The substitution method is a powerful algebraic technique that simplifies the process of solving equations, especially when dealing with more complex polynomial forms. In this method, you temporarily replace a part of the equation with a new variable to reduce complexity, making the equation easier to solve.
In our earlier exercise, we had a quartic equation \(4x^4 + 17x^2 + 4 = 0\). By observing its structure, you can note that it resembles a quadratic format. The substitution method allows us to let \(y = x^2\), transforming the equation into a more recognizable quadratic form \(4y^2 + 17y + 4 = 0\).

• This substitution makes it easier to apply the quadratic formula or other techniques to solve for \(y\).
• Once we solve for \(y\), we back-substitute to find the original variable \(x\).

This method is especially useful when direct computation is challenging, providing a bridge to simpler algebraic forms.

Quadratic Formula

The quadratic formula is an essential tool in algebra used to find the solutions of quadratic equations. It is applicable when equations are in the form \(ax^2 + bx + c = 0\). The formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where:

• \(a,\) \(b,\) and \(c\) are the coefficients of the quadratic equation,
• \(\sqrt{b^2 - 4ac}\) is the discriminant, determining the nature of the roots.

Using the quadratic formula helps to efficiently find the roots (or solutions) of a quadratic equation. In the problem at hand, it helped determine the values of \(y\). Here:

• By substituting \(y\) into our equation, we found two potential values of \(y\),
• However, upon back-substitution, neither constitutes a real solution for \(x\).

The quadratic formula is invaluable, as it provides precise solutions, including ones that may involve irrational or complex numbers. With it, we directly access the tools to explore the possibilities encapsulated within quadratic forms.