Problem 19
Question
\(-\frac{1}{4} x^{2}=-4\)
Step-by-Step Solution
Verified Answer
x = 4 or x = -4
1Step 1: Isolate the Quadratic Term
To isolate the quadratic term, divide both sides of the equation by \(-\frac{1}{4}\). This gives \(x^2=16\).
2Step 2: Take the Square Root of Both Sides
Next, take the square root of both sides to solve for \(x\). Remember to include both the positive and negative roots. \(x = \pm\sqrt{16}\).
3Step 3: Simplify the Square Root
Simplify \sqrt{16}\. Therefore, \(x = \pm 4\).
Key Concepts
Isolating the Quadratic TermTaking the Square RootPositive and Negative Roots
Isolating the Quadratic Term
The first step in solving a quadratic equation like \(-\frac{1}{4} x^{2}=-4\) is to isolate the quadratic term. This means we need to get \({ x^{2}}\) by itself on one side of the equation. In this exercise, you divide both sides of the equation by \(-\frac{1}{4}\). Division helps to eliminate the coefficient (the number attached to \({ x^{2}}\)) from the quadratic term. After dividing, you have: \({ x^{2}=16}\). Now, the term with the variable \({ x}\) is isolated, setting you up to solve for \({ x}\) in the next step.
Taking the Square Root
After isolating the quadratic term, the next step is to take the square root of both sides of the equation. This step allows you to solve for \({ x}\). When taking the square root, always remember there are two possible roots: the positive and the negative. So, from \({ x^{2}=16}\), you get: \({ x = \pm\sqrt{16}}\). The \({ \pm \sqrt{16}}\) indicates both the positive and negative square root values.
Positive and Negative Roots
In the final step, simplify the square root. \({ \sqrt{16}}\) simplifies to \({ 4}\). Hence, you get: \({ x=4}\) and \({ x=-4}\). It is crucial to include both the positive and negative roots because squaring either value (4 or -4) would give you the same original quadratic term \({ x^{2}}\). So, the solutions to the equation \(-\frac{1}{4} x^{2} = -4\) are \({ x=4}\) and \({ x=-4}\). Always double-check your work by substituting the solutions back into the original equation to ensure they satisfy it.