Problem 39 - Algebra 1: Concepts and Skills

Question

Solve the equation algebraically. Check your solutions by graphing. $$x^{2}-4=12$$

Step-by-Step Solution

Answer

The solutions to the equation \(x^{2}-4=12\) are \(x=4\) and \(x=-4\)

Step 1: Simplify the Equation

In the given equation \(x^{2}-4=12\), we can start simplifying by isolating \(x^{2}\) on one side of the equation. We do so by adding 4 to both sides of the equation which gives us: \(x^{2} = 12 + 4\), or \(x^{2} = 16\)

Step 2: Solve for \(x\)

After simplifying the equation in the previous step to \(x^{2}=16\), we can now solve for \(x\). The square root of 16 is \(±4\), therefore, the solution to the equation is \(x = ±4\). This means \(x\) can either be 4 or -4.

Step 3: Check the Solution by Graphing

We can check the correctness of the solution by graphing the function \(y = x^{2}-4\)-12. When \(x = 4\) or \(x = -4\), the value of \(y\) should be zero. We can plot a graph with \(x\) on one axis and \(y\) on the other, and make a curve for the equation. For \(x=4\) or \(x=-4\), the curve should intersect the x-axis, confirming that our solutions are correct.

Key Concepts

Algebraic Solutions

Solving quadratic equations algebraically is a fundamental skill in algebra. To illustrate this, let's look at the equation from our exercise: \(x^2 - 4 = 12\). The goal is to find the value of \(x\) that satisfies this equation.

The first step involves moving all terms to one side to isolate the quadratic term, in this case \(x^2\). By adding 4 to both sides, we set the stage for extracting the square root to find \(x\). The equation simplifies to \(x^2 = 16\), which possess a clear solution once we apply the square root. This is where knowledge of square roots becomes essential, we derive \(x = \pm4\) since both 4 and -4 squared will return 16. It's crucial to include the \(\pm\) sign because it represents both the positive and negative roots of the number.

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• Isolate the quadratic term.
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• Simplify the equation.
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• Extract the square root, accounting for both positive and negative outcomes.

Checking the solution by plotting ensures that the algebraic manipulation aligns with the visual representation of the function on the graph.

Quadratic Functions Graphing

Graphing quadratic functions provides a visual perspective on solutions to equations. The quadratic function from our exercise, \(y = x^2 - 16\), can be graphed to confirm the solutions found algebraically.

A quadratic function is typically represented by a parabola when graphed. In our function, the parabola will be oriented upwards because the coefficient of \(x^2\) is positive, and the vertex of the parabola will be at the point \(x, y) = (0, -16)\), based on the constant term in the equation.

The solutions to the equation, \(x = \pm4\), are referred to as the 'x-intercepts' or 'roots' of the function because they are the points where the parabola crosses the x-axis. When graphing, ensuring accuracy in the plot will affirm that these intercepts occur at \(x = 4\) and \(x = -4\).

Key Points in Graphing:

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• The graph of a quadratic is a parabola.
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• The coefficient of \(x^2\) determines the direction of the parabola.
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• The constant term in the function influences the vertex's location.
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• Solutions (roots) are where the graph intersects the x-axis.

The intersection points validate our algebraic solutions graphically.

Square Roots in Algebra

Understanding square roots is essential when solving quadratic equations. A square root essentially asks the question, 'What number, multiplied by itself, gives the original number?' In the context of our problem, we want to find which number squared equals 16.

The notation for the square root is \(\sqrt{...}\), where the number inside the radical sign is the one we're finding the root of. The square root of 16 is 4, but we must not forget that \(x^2 = 16\) has two solutions: \(x = 4\) and \(x = -4\), because squaring either of these numbers will return 16.

We thus encounter the concept of 'principal' and 'negative' square roots. The principal square root is the non-negative root, represented by \(\sqrt{16} = 4\), while the negative root is addressed by putting a minus sign in front of the square root, \( -\sqrt{16} = -4\). Both roots are equally valid in the realm of real numbers.

Takeaways on Square Roots:

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• A square root finds a number that, when squared, returns the original number.
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• \tThe principal square root is non-negative; the negative square root is also a valid solution for quadratic equations.
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• In our case, \(\sqrt{16}\) results in \(\pm 4\), accounting for both the principal and negative roots.

Grasping this concept ensures accurate solutions to quadratics and provides a foundation for more complex algebraic problems.