Problem 21 - Algebra 1

Question

Solve the equation algebraically. Check the solutions graphically. $$ 2 x^{2}=32 $$

Step-by-Step Solution

Answer

The algebraic solutions for the equation are \(x=4\) and \(x=-4\).

Step 1: Simplify the equation

The given equation is \(2 x^{2}=32\). To isolate \(x^{2}\), divide both sides of the equation by 2. The simplified equation is \(x^{2}=16\).

Step 2: Solve for x

To solve for 'x', take the square root of both sides. Remember that when you take the square root of a value, there is a positive and a negative solution. After taking the square root, you get \(x=4\) and \(x=-4\).

Step 3: Check solutions graphically

On plotting the equation \(y=2x^{2}\), it can be noticed that the parabola intersects the x-axis at the points \(x=4\) and \(x=-4\). This graphically verifies the solution obtained algebraically.

Key Concepts

Factoring

Factoring is a powerful technique used to solve quadratic equations by expressing them as a product of simpler expressions. It is based on finding numbers or expressions that can multiply together to give the original quadratic expression. In the given exercise, the equation is already simplified to the form of a perfect square, such as \( x^2 = 16 \). However, if we had an equation in the standard quadratic form, like \( ax^2 + bx + c = 0 \), factoring would be very useful. Here's how the process works:

• First, look for any common factors in all terms of the quadratic expression.
• Next, try breaking down the quadratic into two binomials that multiply together. These binomials will have the form \((x - p)(x - q) = 0\), where \(p\) and \(q\) are numbers that add up to \(b\) and multiply to \(a \times c\).
• Finally, solve the equation by setting each binomial equal to zero, yielding two potential solutions.

While factoring might seem complicated at first, with practice, it becomes easier to spot patterns and quickly solve quadratic equations. This technique is especially useful when the equation can be easily broken down, offering a clear pathway to find the roots of the quadratic.

Square Roots

Taking square roots is one of the most direct methods to solve quadratic equations, particularly when the equation can be easily rearranged into the form \( x^2 = d \). This was the approach taken in the original exercise step by step solution.

• To begin solving, you isolate the expression \( x^2 \) on one side of the equation as much as possible. The goal is to have a simple expression that only includes \( x^2 \) on one side.
• Once this is done, take the square root of both sides. It's important to remember that the square root operation introduces two possible solutions: one positive and one negative. This is because both \( (\sqrt{d})^2 \) and \( (-\sqrt{d})^2 \) will equal \( d \).
• The expression \( x^2 = 16 \) simplifies to \( x = \pm 4 \), which means the solutions are \( x = 4 \) and \( x = -4 \).

This method is particularly useful for equations like the one in the exercise, where the value can be easily reduced to a perfect square, making it quick and efficient to find solutions.

Graphing Parabolas

Graphing parabolas is an excellent way to visually confirm solutions to quadratic equations. A parabola is the graphical representation of a quadratic function and typically has the form \( y = ax^2 + bx + c \). In the case of the exercise, you are graphing the equation \( y = 2x^2 \).

• This equation represents a parabola that opens upwards, with its vertex at the origin \((0, 0)\) since there is no \( b \) or \( c \) term.
• To graph the parabola, plot several points by choosing values of \( x \) and calculating the corresponding \( y \) values. For example, if \( x = 1 \), then \( y = 2(1)^2 = 2 \).
• Draw the curve smoothly through these points to create the parabola. This visual representation allows you to see where the parabola intersects the \( x \)-axis, confirming the roots found algebraically. In this case, intersections at \( x = 4 \) and \( x = -4 \) verify the solutions of the equation \( x^2 = 16 \).

Using graphing as a cross-check not only confirms the solutions but also reinforces understanding of the behavior of quadratic functions based on their coefficients.