Problem 68 - Algebra 1

Question

Solve the quadratic equation by finding square roots or by using the quadratic formula. Explain why you chose the method. $$5 x^{2}=25$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation \(5x^2 = 25\) are \(x = \sqrt {5}\) or \(x = -\sqrt {5}\), which can also be represented as \(x = - 2.236067977\) and \(x = 2.236067977\).

Step 1: Understand the equation

The equation given is \(5x^2 = 25\). This is a quadratic equation because it is of the form \(ax^2 = c\), where \(a\) and \(c\) are numbers.

Step 2: Isolate x^2

To isolate \(x^2\), divide both sides of the equation by 5. Then you will get \(x^2 = 5\). This isolates \(x^2\) by itself on one side of the equation.

Step 3: Use the square root property

Next, you can find \(x\) by finding the square root of both sides of the equation. However, when square rooting, one must consider both the positive and negative roots. This will yield \(x = \sqrt {5}\) or \(x = -\sqrt {5}\).

Step 4: Rationalize the denominator (Optional)

The solutions can be left as \(x = -\sqrt {5}\) and \(x = \sqrt {5}\). But in case if the answer needs to be provided in simplified radical form or decimal form, rationalize the denominator which gives two possible solutions for this quadratic equation: \(x = - 2.236067977\) and \(x = 2.236067977\) .

Key Concepts

Solving Quadratics

Quadratic equations appear in the form \(ax^2 + bx + c = 0\). Solving these equations means finding the value of \(x\) that makes the equation true. There are various methods to solve quadratics:

• Factoring: Breaking down the equation into factors.
• Completing the square: Rearranging to achieve a perfect square trinomial.
• Graphing: Finding the points where the graph crosses the x-axis.
• Square Root Property: In cases like \(ax^2 = c\), finding \(x\) by taking square roots.
• Quadratic Formula: A universal method that works with all quadratic equations.

Choosing a method depends on the form of the equation and what's most comfortable. In this exercise, using the square root property was efficient because the equation is already set up as \(ax^2 = c\). If the equation had a non-zero linear term \(bx\), other methods might have been more suitable.

Square Root Property

The square root property is useful when you have a simple equation like \(ax^2 = c\). To solve it:
1. First, ensure the \(x^2\) term is isolated on one side of the equation. For example, divide both sides by \(a\) if necessary.
2. Once isolated, take the square root of both sides of the equation. Remember, taking squares introduces both positive and negative roots.
For the equation \(x^2 = 5\), applying the square root property yields two solutions: \(x = \sqrt{5}\) and \(x = -\sqrt{5}\).
This method is straightforward and quick when the quadratic equation is in a simplified form. It's critical to remember that whenever you take the square root of both sides of an equation, you should consider both the positive and negative solutions.

Quadratic Formula

The quadratic formula offers a reliable way to find solutions for any quadratic equation. It uses the formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]
Here's why the quadratic formula is handy:

• Works on any polynomial of degree 2, even when other methods fail.
• Always provides two solutions based on the discriminant \(b^2-4ac\).
• If \(b^2-4ac\) is positive, there are two distinct real roots. If zero, one real root. If negative, complex roots occur.

For equations lacking a linear term, like our exercise \(5x^2 = 25\), the quadratic formula is not necessary. However, it's a crucial tool when equations are more complex and need a comprehensive solution approach.