Problem 54 - Elementary Algebra

Question

For the following problems, solve the equations, if possible. $$ -7 b^{2}=-63 $$

Step-by-Step Solution

Answer

Question: Solve the equation -7b^2 = -63 for the variable b. Answer: b = -3, b = 3

Step 1: Set up the given equation

We are given the equation: $$ -7 b^{2}=-63 $$

Step 2: Isolating \(b^{2}\) term

To isolate the \(b^{2}\) term, we should divide both sides of the equation by \(-7\): $$ \frac{-7 b^{2}}{-7} = \frac{-63}{-7} $$ This simplifies to: $$ b^{2} = 9 $$

Step 3: Solving for \(b\)

Now that we have \(b^{2}\) isolated, we can solve for \(b\) by taking the square root of both sides: $$ \sqrt{b^{2}} = \pm \sqrt{9} $$ So, the solutions for \(b\) are: $$ b = -3, \quad b = 3 $$

Key Concepts

Isolating the Variable

Solving quadratic equations often starts with the fundamental step of isolating the variable. This means rearranging the equation such that the term containing the variable to be solved for is on one side of the equation, and everything else is on the other side. This makes it easier to apply further algebraic techniques to solve the equation.

For the given problem, we start with \[ -7b^2 = -63 \]. To isolate the variable term, which is \(b^2\), we need to get rid of the coefficient \( -7 \) that is attached to it. We do this by dividing both sides of the equation by \( -7 \), leaving us with \( b^2 = 9 \).

This step is crucial, as it sets the stage for employing other methods, like taking square roots, to find the actual solutions for the variable. Remember that whatever operation you perform on one side of the equation, you must do the same to the other side to keep the equation balanced.

Square Roots

The concept of square roots is integral to solving quadratic equations where the variable is squared. A square root essentially asks the question: what number, when multiplied by itself, gives the given number? For instance, if we have \(b^2 = 9\), we ask what number squared equals 9.

To find \(b\), we take the square root of both sides of the equation, which gives us \(\sqrt{b^2} = \pm\sqrt{9}\). Why the \(\pm\)? Because both positive and negative values of a number, when squared, will yield the same positive result. Here, \(\sqrt{9}\) equals both 3 and -3, so we get two solutions for \(b\): \(b = 3\) and \(b = -3\).

It's important to include both the positive and negative solutions unless the context of the problem restricts the answer to only one of them. The principal square root is always positive, but when solving equations, we must consider both the principle (positive) and the negative square roots.

Quadratic Formula

The quadratic formula provides a straightforward method to solve any quadratic equation in the form \(ax^2 + bx + c = 0\). The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It is especially useful when the quadratic equation is not easily factorable or when dealing with coefficients that are not perfect squares.

While the quadratic formula was not necessary for the simplicity of the given exercise, it is important to understand its usefulness. Had the original equation been a more complex quadratic, the quadratic formula could have been applied after properly setting the equation to the standard form. The \(\pm\) symbol in the formula represents the two possible solutions for \(x\), resulting from the fact that the square root operation can yield both positive and negative numbers.

Knowing when to apply the quadratic formula is as crucial as understanding how to derive it from completing the square or how to efficiently compute the discriminant (\(b^2 - 4ac\)), which determines the nature and number of the roots.