To find the solutions to the equation [tex]\(\frac{1}{4} x^2 = -\frac{1}{2} x + 2\)[/tex], let's break down the steps needed to solve it.
First, we need to rearrange this equation so we have all terms on one side, forming a standard quadratic equation. Here's the original equation:
[tex]\[
\frac{1}{4} x^2 = -\frac{1}{2} x + 2
\][/tex]
Let's get rid of the fraction by multiplying every term by 4 to clear the denominators. This simplifies our equation:
[tex]\[
x^2 = -2x + 8
\][/tex]
Next, move all terms to one side of the equation to set it to 0:
[tex]\[
x^2 + 2x - 8 = 0
\][/tex]
Now, we have a standard quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex].
To solve for [tex]\(x\)[/tex], we can use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For our equation [tex]\(x^2 + 2x - 8 = 0\)[/tex], the coefficients are [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -8\)[/tex].
Plug these values into the quadratic formula:
[tex]\[
x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{-2 \pm \sqrt{4 + 32}}{2}
\][/tex]
[tex]\[
x = \frac{-2 \pm \sqrt{36}}{2}
\][/tex]
Since [tex]\(\sqrt{36} = 6\)[/tex], we get:
[tex]\[
x = \frac{-2 \pm 6}{2}
\][/tex]
This results in two solutions:
[tex]\[
x = \frac{-2 + 6}{2} = \frac{4}{2} = 2
\][/tex]
[tex]\[
x = \frac{-2 - 6}{2} = \frac{-8}{2} = -4
\][/tex]
So, the solutions are [tex]\(x = -4\)[/tex] and [tex]\(x = 2\)[/tex].
Let's match these solutions with the given choices:
1. [tex]\(-4\)[/tex] and [tex]\(2\)[/tex]
2. [tex]\(-4\)[/tex] and [tex]\(1\)[/tex]
3. [tex]\(0\)[/tex] and [tex]\(4\)[/tex]
4. [tex]\(1\)[/tex] and [tex]\(4\)[/tex]
The correct choice is:
[tex]\(-4\)[/tex] and [tex]\(2\)[/tex]
