Question
What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
x = StartFraction negative 5 Over 2 EndFraction – StartFraction StartRoot 31 EndRoot Over 2 EndFraction and x = StartFraction negative 5 Over 2 EndFraction + StartFraction StartRoot 31 EndRoot Over 2 EndFraction
x = StartFraction negative 5 Over 2 EndFraction – StartFraction StartRoot 37 EndRoot Over 8 EndFraction and x = StartFraction negative 5 Over 2 EndFraction + StartFraction StartRoot 37 EndRoot Over 8 EndFraction
x = StartFraction 5 Over 2 EndFraction – StartFraction StartRoot 31 EndRoot Over 2 EndFraction and x = StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 31 EndRoot Over 2 EndFraction
x = StartFraction 5 Over 2 EndFraction – StartFraction StartRoot 37 EndRoot Over 8 EndFraction and x = StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 37 EndRoot Over 8 EndFraction
x = StartFraction negative 5 Over 2 EndFraction – StartFraction StartRoot 31 EndRoot Over 2 EndFraction and x = StartFraction negative 5 Over 2 EndFraction + StartFraction StartRoot 31 EndRoot Over 2 EndFraction
x = StartFraction negative 5 Over 2 EndFraction – StartFraction StartRoot 37 EndRoot Over 8 EndFraction and x = StartFraction negative 5 Over 2 EndFraction + StartFraction StartRoot 37 EndRoot Over 8 EndFraction
x = StartFraction 5 Over 2 EndFraction – StartFraction StartRoot 31 EndRoot Over 2 EndFraction and x = StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 31 EndRoot Over 2 EndFraction
x = StartFraction 5 Over 2 EndFraction – StartFraction StartRoot 37 EndRoot Over 8 EndFraction and x = StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 37 EndRoot Over 8 EndFraction
Answers
Answered by
GPT-4o mini
To find the zeros of the quadratic function \( f(x) = 2x^2 - 10x - 3 \), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = -10 \), and \( c = -3 \).
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = (-10)^2 = 100
\]
\[
4ac = 4 \cdot 2 \cdot (-3) = -24
\]
\[
b^2 - 4ac = 100 - (-24) = 100 + 24 = 124
\]
Now we can substitute these values into the quadratic formula:
\[
x = \frac{-(-10) \pm \sqrt{124}}{2 \cdot 2}
\]
\[
x = \frac{10 \pm \sqrt{124}}{4}
\]
Next, we simplify \( \sqrt{124} \):
\(\sqrt{124} = \sqrt{4 \cdot 31} = 2\sqrt{31}\)
Now substituting back in:
\[
x = \frac{10 \pm 2\sqrt{31}}{4}
\]
We can simplify this expression as follows:
\[
x = \frac{10}{4} \pm \frac{2\sqrt{31}}{4}
\]
\[
x = \frac{5}{2} \pm \frac{\sqrt{31}}{2}
\]
Thus, the zeros of the quadratic function \( f(x) = 2x^2 - 10x - 3 \) are:
\[
x = \frac{5}{2} - \frac{\sqrt{31}}{2} \quad \text{and} \quad x = \frac{5}{2} + \frac{\sqrt{31}}{2}
\]
Among the provided options, the correct answer is:
**x = \( \frac{5}{2} - \frac{\sqrt{31}}{2} \) and x = \( \frac{5}{2} + \frac{\sqrt{31}}{2} \)**.
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = -10 \), and \( c = -3 \).
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = (-10)^2 = 100
\]
\[
4ac = 4 \cdot 2 \cdot (-3) = -24
\]
\[
b^2 - 4ac = 100 - (-24) = 100 + 24 = 124
\]
Now we can substitute these values into the quadratic formula:
\[
x = \frac{-(-10) \pm \sqrt{124}}{2 \cdot 2}
\]
\[
x = \frac{10 \pm \sqrt{124}}{4}
\]
Next, we simplify \( \sqrt{124} \):
\(\sqrt{124} = \sqrt{4 \cdot 31} = 2\sqrt{31}\)
Now substituting back in:
\[
x = \frac{10 \pm 2\sqrt{31}}{4}
\]
We can simplify this expression as follows:
\[
x = \frac{10}{4} \pm \frac{2\sqrt{31}}{4}
\]
\[
x = \frac{5}{2} \pm \frac{\sqrt{31}}{2}
\]
Thus, the zeros of the quadratic function \( f(x) = 2x^2 - 10x - 3 \) are:
\[
x = \frac{5}{2} - \frac{\sqrt{31}}{2} \quad \text{and} \quad x = \frac{5}{2} + \frac{\sqrt{31}}{2}
\]
Among the provided options, the correct answer is:
**x = \( \frac{5}{2} - \frac{\sqrt{31}}{2} \) and x = \( \frac{5}{2} + \frac{\sqrt{31}}{2} \)**.