Problem 60 - Elementary Algebra

Question

For the following problems, solve each of the quadratic equations using the method of extraction of roots. $$ (x+10)^{2}=c^{2}, \text { for } x $$

Step-by-Step Solution

Answer

Answer: The two possible solutions for x are: x = -10-c and x = -10+c.

Step 1: Rewriting the equation

Rewrite the given quadratic equation as an equation involving x only: $$(x+10)^2 - c^2 = 0$$

Step 2: Applying the difference of squares formula

We can notice that this equation is the difference of squares. So, apply the difference of squares formula here: $$(x+10+c)((x+10)-c) = 0$$

Step 3: Solving for x

Now, we have two equations to solve for x: 1) \(x+10+c = 0\) 2) \(x+10-c = 0\) For the first equation, rearrange the terms to isolate x: \(x = -10-c\) For the second equation, rearrange the terms to isolate x: \(x = -10+c\)

Step 4: Final solution

As we have two possible solutions for x: 1) \(x = -10-c\) 2)\(x = -10+c\) Therefore, the quadratic equation \((x+10)^2 = c^2\) can be solved using the method of extraction of roots, resulting in two possible solutions for x.

Key Concepts

Extraction of Roots

The extraction of roots is a method used to solve quadratic equations that involve a perfect square. It revolves around the concept of reverse engineering the squaring process. To apply this method, you typically start by setting up the equation in the form \(a^2 = b\), where \(a\) is an expression containing the variable, and \(b\) is a constant or another expression. Once in this form, you can extract the square root of both sides, which gives you \(a = \pm\sqrt{b}\). This \(\pm\) symbol represents the concept that there are often two solutions to a quadratic equation: one positive and one negative. In our exercise, you'll notice that \(c^2\) is the squared term, giving us a clear path to use the extraction of roots.

Difference of Squares Formula

The difference of squares is an algebraic pattern that emerges when you subtract one squared number from another, represented by the formula \(a^2 - b^2 = (a + b)(a - b)\). It is a useful factoring technique that breaks down the original expression into the product of a sum and a difference, which can then be solved as separate linear equations. In the problem we're exploring, we use this pattern to transform \( (x+10)^2 - c^2\) into a difference of squares, allowing us to factor it into \( (x + 10 + c)(x + 10 - c)\). This step is critical because it sets the stage for the quadratic equation to be broken into simpler linear equations, which can be independently solved to find values of \(x\).

Quadratic Equation Solutions

Quadratic equations are second-degree polynomial equations and generally have the form \(ax^2 + bx + c = 0\). These equations can have up to two solutions. The solutions can be real or complex numbers depending on the discriminant \(b^2 - 4ac\). If the discriminant is positive, there are two distinct real number solutions. If it is zero, there is one real number solution. If the discriminant is negative, the solutions are complex numbers. In our example, the original equation \( (x+10)^2 = c^2\) is a quadratic in disguise, and after applying the difference of squares, we end up with two linear equations. Each of these linear equations provides a potential solution to the original quadratic equation.

Isolate Variable

To isolate the variable means to rearrange the algebraic equation in such a way that the variable we're solving for is on one side of the equation all by itself. This process often involves adding, subtracting, multiplying, or dividing both sides of the equation by the same number. In our steps, we worked with the two linear factors \(x+10+c = 0\) and \(x+10-c = 0\), which arose from using the difference of squares formula. Our goal was to find the value of \(x\) that satisfies each equation. To do that, we rearranged each equation to solve for \(x\), yielding two possible solutions, \(x = -10-c\) and \(x = -10+c\). Understanding how to isolate the variable is essential in algebra, as it's a fundamental skill used in solving equations.