Which is not occur finding zero The image presents a mathematical problem involving the polynomial function $ f(x) = 4x^2 - 8x - 296 $. The task is to identify the step where an error first occurs when finding the zeros of $ f(x) $ using the quadratic formula.### Quadratic Formula Steps:1. **Step 1:** \[ x = \frac{-8 \pm \sqrt{(-8)^2 - 4(4)(-296)}}{2(4)} \] 2. **Step 2:** \[ x = \frac{-8 \pm \sqrt{64 + 4736}}{8} \]3. **Step 3:** \[ x = \frac{-8 \pm \sqrt{4800}}{8} \]4. **Step 4:** \[ x = \frac{-8 \pm \sqrt{1600 \cdot 3}}{8} \]5. **Step 5:** \[ x = \frac{-8 \pm \frac{40 \sqrt{3}}{8}}{8} \]6. **Step 6:** \[ x = -8 \pm 5\sqrt{3} \]### Task:- **(a)** Identify the initial step where an error occurs in calculating the zeros of $ f(x) $.### Final Question:- Find the zeros of $ f(x) $ in exact form. Separate multiple answers with commas. If there’s no error in part (a), answer as "DNE" (Does Not Exist).**Note:** There is a visible error in Step 6 where the simplification shows incorrect substitution or calculation based on the previous steps.

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Description: Which is not occur finding zero The image presents a mathematical problem involving the polynomial function $ f(x) = 4x^2 - 8x - 296 $. The task is to identify the step where an error first occurs when finding the zeros of $ f(x) $ using the quad

We compare the equation f(x)=4x2 -8x - 296 with y=ax2+bx+c So, a=4, b=-8 and c= -296 Using the…

Answered: Suppose f(x) = 4x2 8x-296. %3D (a) In…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Which is not occur finding zero

The image presents a mathematical problem involving the polynomial function $ f(x) = 4x^2 - 8x - 296 $. The task is to identify the step where an error first occurs when finding the zeros of $ f(x) $ using the quadratic formula.

### Quadratic Formula Steps:

1. **Step 1:**
\[
x = \frac{-8 \pm \sqrt{(-8)^2 - 4(4)(-296)}}{2(4)}
\]

2. **Step 2:**
\[
x = \frac{-8 \pm \sqrt{64 + 4736}}{8}
\]

3. **Step 3:**
\[
x = \frac{-8 \pm \sqrt{4800}}{8}
\]

4. **Step 4:**
\[
x = \frac{-8 \pm \sqrt{1600 \cdot 3}}{8}
\]

5. **Step 5:**
\[
x = \frac{-8 \pm \frac{40 \sqrt{3}}{8}}{8}
\]

6. **Step 6:**
\[
x = -8 \pm 5\sqrt{3}
\]

### Task:
- **(a)** Identify the initial step where an error occurs in calculating the zeros of $ f(x) $.

### Final Question:
- Find the zeros of $ f(x) $ in exact form. Separate multiple answers with commas. If there’s no error in part (a), answer as "DNE" (Does Not Exist).

**Note:** There is a visible error in Step 6 where the simplification shows incorrect substitution or calculation based on the previous steps.

Expert Solution

Step 1

We compare the equation f(x)=4x² -8x - 296 with y=ax²+bx+c

So, a=4, b=-8 and c= -296

Using the quadratic formula we get:

$x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} x = \frac{- (- 8) \pm \sqrt{{(- 8)}^{2} - 4 (4) (- 296)}}{2 (4)} x = \frac{8 \pm \sqrt{64 + 4736}}{8} x = \frac{8 \pm \sqrt{4800}}{8} x = \frac{8 \pm 40 \sqrt{3}}{8} x = \frac{8 (1 \pm 5 \sqrt{3})}{8} x = 1 \pm 5 \sqrt{3}$

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