Question
Solve the equation by graphing the related function.one halfx Superscript 2minus2equals0Question content area bottomPart 1Select the correct choice below and, if necessary, fill in the answer box to complete your choice.A.xequalsenter your response here (Simplify your answer. Use a comma to separate answers as needed.)B.The solution is not a real number.
Answers
GPT-4o mini
To solve the equation \(\frac{1}{2}x^2 - 2 = 0\) by graphing the related function, we can first rewrite the equation as a function:
\[ f(x) = \frac{1}{2}x^2 - 2 \]
Next, we need to find the x-values where this function is equal to zero, since that corresponds to the solutions of the equation.
1. **Graph the function**:
The function \(f(x)\) is a quadratic function that opens upwards (since the coefficient of \(x^2\) is positive). It will have its vertex at the minimum point, which can be found using the vertex formula \(x = -\frac{b}{2a}\), where \(f(x) = ax^2 + bx + c\). In our case, \(a = \frac{1}{2}\), \(b = 0\), and \(c = -2\).
\[
x = -\frac{0}{2 \cdot \frac{1}{2}} = 0
\]
Plugging \(x = 0\) into the function:
\[
f(0) = \frac{1}{2}(0)^2 - 2 = -2
\]
This means the vertex is at the point \((0, -2)\).
2. **Find the x-intercepts**:
To find the x-intercepts, we set \(f(x)\) to zero:
\[
\frac{1}{2}x^2 - 2 = 0
\]
Multiplying both sides by 2 to eliminate the fraction:
\[
x^2 - 4 = 0
\]
Factoring the equation:
\[
(x - 2)(x + 2) = 0
\]
This gives us two solutions:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
Thus, the solutions to the equation \(\frac{1}{2}x^2 - 2 = 0\) are:
\[
x = 2, -2
\]
### Final Choice:
A. \(x = -2, 2\) (Simplified answer).
\[ f(x) = \frac{1}{2}x^2 - 2 \]
Next, we need to find the x-values where this function is equal to zero, since that corresponds to the solutions of the equation.
1. **Graph the function**:
The function \(f(x)\) is a quadratic function that opens upwards (since the coefficient of \(x^2\) is positive). It will have its vertex at the minimum point, which can be found using the vertex formula \(x = -\frac{b}{2a}\), where \(f(x) = ax^2 + bx + c\). In our case, \(a = \frac{1}{2}\), \(b = 0\), and \(c = -2\).
\[
x = -\frac{0}{2 \cdot \frac{1}{2}} = 0
\]
Plugging \(x = 0\) into the function:
\[
f(0) = \frac{1}{2}(0)^2 - 2 = -2
\]
This means the vertex is at the point \((0, -2)\).
2. **Find the x-intercepts**:
To find the x-intercepts, we set \(f(x)\) to zero:
\[
\frac{1}{2}x^2 - 2 = 0
\]
Multiplying both sides by 2 to eliminate the fraction:
\[
x^2 - 4 = 0
\]
Factoring the equation:
\[
(x - 2)(x + 2) = 0
\]
This gives us two solutions:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
Thus, the solutions to the equation \(\frac{1}{2}x^2 - 2 = 0\) are:
\[
x = 2, -2
\]
### Final Choice:
A. \(x = -2, 2\) (Simplified answer).