Which Is A Zero Of The Quadratic Function $f(x) = 16x^2 + 32x - 9$?A. $x = -5.25$ B. $x = -2.25$ C. $x = -1.25$ D. $x = -0.25$

Introduction

Quadratic equations are a fundamental concept in mathematics, and they have numerous applications in various fields such as physics, engineering, and economics. A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ is the variable. In this article, we will focus on finding the zeroes of a quadratic function, which is a crucial concept in solving quadratic equations.

What are Zeroes of a Quadratic Function?

The zeroes of a quadratic function are the values of $x$ that make the function equal to zero. In other words, they are the solutions to the equation $f(x) = 0$. Finding the zeroes of a quadratic function is essential in understanding the behavior of the function, and it has numerous applications in various fields.

The Quadratic Formula

The quadratic formula is a powerful tool for finding the zeroes of a quadratic function. The quadratic formula states that for a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

This formula is derived from the fact that the quadratic equation can be factored as $(x - r)(x - s) = 0$, where $r$ and $s$ are the zeroes of the function.

Applying the Quadratic Formula

To find the zeroes of the quadratic function $f(x) = 16x^2 + 32x - 9$, we can use the quadratic formula. Plugging in the values of $a$, $b$, and $c$, we get:

$x = \frac{-32 \pm \sqrt{32^2 - 4(16)(-9)}}{2(16)}$

Simplifying the expression, we get:

$x = \frac{-32 \pm \sqrt{1024 + 576}}{32}$

$x = \frac{-32 \pm \sqrt{1600}}{32}$

$x = \frac{-32 \pm 40}{32}$

This gives us two possible solutions:

$x = \frac{-32 + 40}{32}$

$x = \frac{8}{32}$

$x = \frac{1}{4}$

$x = \frac{-32 - 40}{32}$

$x = \frac{-72}{32}$

$x = -\frac{9}{4}$

Evaluating the Solutions

Now that we have found the two possible solutions, we need to evaluate them to determine which one is correct. We can do this by plugging the solutions back into the original equation.

For $x = \frac{1}{4}$:

$f(\frac{1}{4}) = 16(\frac{1}{4})^2 + 32(\frac{1}{4}) - 9$

$f(\frac{1}{4}) = 16(\frac{1}{16}) + 8 - 9$

$f(\frac{1}{4}) = 1 + 8 - 9$

$f(\frac{1}{4}) = 0$

Since $f(\frac{1}{4}) = 0$, we can conclude that $x = \frac{1}{4}$ is a zero of the quadratic function.

For $x = -\frac{9}{4}$:

$f(-\frac{9}{4}) = 16(-\frac{9}{4})^2 + 32(-\frac{9}{4}) - 9$

$f(-\frac{9}{4}) = 16(\frac{81}{16}) - 72 - 9$

$f(-\frac{9}{4}) = 81 - 72 - 9$

$f(-\frac{9}{4}) = 0$

Since $f(-\frac{9}{4}) = 0$, we can conclude that $x = -\frac{9}{4}$ is also a zero of the quadratic function.

Conclusion

In this article, we have discussed the concept of zeroes of a quadratic function and how to find them using the quadratic formula. We have applied the quadratic formula to the quadratic function $f(x) = 16x^2 + 32x - 9$ and found two possible solutions. We have then evaluated the solutions to determine which one is correct. We have concluded that both $x = \frac{1}{4}$ and $x = -\frac{9}{4}$ are zeroes of the quadratic function.

Final Answer

Based on our analysis, we can conclude that the correct answer is:

• A. $x = -5.25$ is incorrect
• B. $x = -2.25$ is incorrect
• C. $x = -1.25$ is incorrect
• D. $x = -0.25$ is incorrect
• The correct answer is $x = -\frac{9}{4}$