rest Question 6 of 16 This test: 16 point(s) possible Thils question: 1 point(s) possible Find the zeros, if any, of the quadratic function using the quadratic formula. What are the $$ x $$-intercepts, if any, of the graph of the function? $$ f(x)=8 x^{2}-3+4 x $$ Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed) A. The zeros and the $$ x $$-intercepts are the same. They are $$ \square $$ B. The zeros and the $$ x $$-intercepts are different. The zeros are $$ \square $$ , the $$ x $$-intercepts are $$ \square $$ C. There is no real zero solution and no $$ x $$-intercept.

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UpStudy AI · Accepted Answer

To find the zeros of the quadratic function $$ f(x) = 8x^2 + 4x - 3 $$, we will use the quadratic formula:

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

where $$ a = 8 $$, $$ b = 4 $$, and $$ c = -3 $$.

### Step 1: Calculate the discriminant

First, we need to calculate the discriminant $$ D = b^2 - 4ac $$:

$$
D = 4^2 - 4 \cdot 8 \cdot (-3)
$$

### Step 2: Substitute values and simplify

Now, let's calculate $$ D $$:

$$
D = 16 + 96 = 112
$$

### Step 3: Apply the quadratic formula

Since the discriminant is positive, we will have two real solutions. Now we can substitute $$ a $$, $$ b $$, and $$ D $$ into the quadratic formula:

$$
x = \frac{-4 \pm \sqrt{112}}{2 \cdot 8}
$$

### Step 4: Simplify the expression

Now, let's simplify $$ \sqrt{112} $$:

$$
\sqrt{112} = \sqrt{16 \cdot 7} = 4\sqrt{7}
$$

Now substituting back into the formula:

$$
x = \frac{-4 \pm 4\sqrt{7}}{16}
$$

This simplifies to:

$$
x = \frac{-1 \pm \sqrt{7}}{4}
$$

### Step 5: Write the final answers

Thus, the zeros of the function are:

$$
x_1 = \frac{-1 + \sqrt{7}}{4}, \quad x_2 = \frac{-1 - \sqrt{7}}{4}
$$

### Conclusion

Since the zeros of the function correspond to the $$ x $$-intercepts of the graph, we can conclude:

A. The zeros and the $$ x $$-intercepts are the same. They are $$ \frac{-1 + \sqrt{7}}{4}, \frac{-1 - \sqrt{7}}{4} $$.
The zeros and the $$ x $$-intercepts are the same. They are $$ \frac{-1 + \sqrt{7}}{4}, \frac{-1 - \sqrt{7}}{4} $$.

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