Problem 34 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation by completing the square. $$x^{2}-6 x=7$$

Step-by-Step Solution

Answer

The solutions to the equation \(x^{2}-6 x=7\) are \(x = -1\) and \(x = 7\).

Step 1: Re-arrange the equation

Rearrange the equation in the form of \(x^{2}-6x - 7=0\). This gives us a clear view of the numbers involved in the equation, making it easier to proceed with other steps.

Step 2: Calculate the number to complete the square

To complete the square we need to add and subtract the square of half of the coefficient of x inside the brackets. That is, \(\left(\frac{-6}{2}\right)^{2} = 9\) which will be added and subtracted within the brackets to complete the square. The equation becomes: \((x^{2}-6x + 9) - 9 - 7=0\).

Step 3: Simplify the equation

Simplify the equation to a square form: \((x-3)^{2} - 16 = 0\). You know that \(a^2 – b^2\) can also be expressed as \((a - b)(a + b)\) which will help solve the equation faster.

Step 4: Solve the equation

By resetting the equation \((x-3)^{2} = 16\), then solving the equation further by taking square root on both sides of the equation we get \(x-3 = ±4\). Finally, by adding 3 to both sides of the equation we get the roots \(x= -1,7\).

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations by transforming them into a perfect square trinomial. This technique helps in converting the quadratic equation into a form that is easier to solve, often enabling the isolation of the variable.

• The first step is to rearrange the equation such that all terms involving the variable are on one side. For example, starting with the equation \(x^2 - 6x = 7\), we rearrange it to \(x^2 - 6x - 7 = 0\) to frame the problem more clearly.
• Next, the main goal is to manipulate the expression so that it becomes a square binomial. Divide the coefficient of \(x\) by 2 and square the result; in this case, \(\left( \frac{-6}{2} \right)^2 = 9\). Add and subtract this square inside the equation, effectively completing the square.
• Now, the equation transforms into \( (x - 3)^2 - 16 = 0 \).

This structure makes it evident how the equation forms a perfect square on one side, with a constant on the other.

Solving Quadratic Equations

Once you have a perfect square trinomial, solving the quadratic equation becomes straightforward. The aim is always to isolate \(x\).

• After completing the square, the equation becomes \((x-3)^2 - 16 = 0\). This expression is in a form that relates directly to the equation \((a)^2 - (b)^2 = 0\), making it approachable to solve further.
• Set the square equal to the constant on the other side by adding 16 to both sides: \((x-3)^2 = 16\).
• Solve for \(x\) by removing the square through taking the square root of both sides: \(x - 3 = ±4\). Remember, anytime you take the square root, consider both positive and negative solutions.
• The final step is to solve these two linear equations: \(x - 3 = 4\) and \(x - 3 = -4\), which gives the solutions \(x = 7\) and \(x = -1\).

This method ensures an accurate solution by carefully working through each logical step.

Quadratic Formula

The quadratic formula is another powerful tool for solving quadratic equations, particularly when completing the square or factoring is complex or difficult. It provides a direct solution to the quadratic equation \(ax^2 + bx + c = 0\) through the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• This formula is derived from the process of completing the square on a general quadratic equation and offers a universal solution for any quadratic equation, given that the equation is in standard form.
• The key component, the discriminant \(b^2 - 4ac\), tells us a lot about the solutions: if it's positive, there are two distinct real solutions; if zero, there's exactly one real solution; and if negative, the equation has no real solutions, only complex ones.
• While completing the square is ideal for equations like \(x^2 - 6x = 7\) due to its symmetry, the quadratic formula is highly effective for less straightforward equations.

By applying these concepts, solving quadratic equations becomes much more accessible.