Problem 81 - Algebra 1: Concepts and Skills

Question

Copy and complete the proof of the following statement: If \(\frac{x^{2}}{4}=9,\) then \(x=\pm 6\)

Step-by-Step Solution

Answer

The solution to the equation \(\frac{x^{2}}{4}=9\) is \(x=\pm 6\)

Step 1: Isolate the term with the variable

The first step is to isolate 'x'. Start from \(\frac{x^{2}}{4}=9\). In order to isolate 'x', multiply both sides by 4. You get \(x^{2}= 4*9\). Simplify this to get \(x^{2}=36\)

Step 2: Apply the square root

Now you have \(x^{2}=36\). The next step is to apply the square root on both sides in order to get rid of the 2nd power on 'x'. Remember, the square root of a square number results in two values: positive and negative. So you get \(x= \pm \sqrt{36}\)

Step 3: Simplify the value

After applying the square root, we simplify to get the final value. So, \(x= \pm \sqrt{36}\) simplifies to \(x= \pm 6\)

Key Concepts

Isolate the Variable

Isolating the variable is a foundational technique in algebra, often used as the first step in solving equations. The goal is to rearrange the equation so that the variable you're solving for is on one side of the equation by itself.

For instance, take the equation from the exercise \( \frac{x^{2}}{4} = 9 \). Here, our aim is to solve for \( x \). To do this effectively, we need to get \( x^{2} \) on its own. This is achieved by performing the same operation on both sides of the equation, ensuring we maintain the equality. In the exercise, we multiplied both sides by 4, turning the left side into \( x^{2}\) and the right side into \( 4 \times 9 \) or \( 36 \). This step simplifies the equation to \( x^{2} = 36 \), successfully isolating \( x^{2} \) and setting up for the next phase, which involves dealing with the square root.

Square Roots

Square roots are fundamental in mathematics, as they allow us to find a value that, when multiplied by itself, gives the original number. In the context of the given exercise, once we have the isolated variable squared, as in \( x^{2} = 36 \), we apply the square root to both sides of the equation to solve for \( x \).

Doing so, we derive \( x = \pm \sqrt{36} \). The plus/minus symbol (\pm) is crucial because squaring either positive or negative numbers yields a positive result. Therefore, when taking the square root of a number, there are generally two possible solutions: one positive and one negative. Hence, the square root of 36 gives us \( x = \pm 6 \), which corresponds to the two potential values for \( x \).

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratics, which are equations in the form of \( ax^{2} + bx + c = 0 \). However, in our exercise, the quadratic formula isn't necessary because the equation is already in a solvable form after isolating the variable and applying square roots.

But for more complex equations, where isolating the variable directly isn't as straightforward, the quadratic formula given by \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \) comes into play. It takes into account all coefficients of the quadratic equation and provides a direct solution for \( x \). Although this method wasn't used in the current example, it's essential to understand as a key approach to solving quadratic equations that cannot be simplified easily through isolation or square roots.