Problem 55 - College Algebra Essentials

Question

By making an appropriate substitution. $$(x-5)^{2}-4(x-5)-21=0$$

Step-by-Step Solution

Answer

Therefore, the solutions of the given quadratic equation are \( x = 12 \) and \( x = 2 \)

Step 1: Define a substitution

Let \( t = x - 5 \). Therefore the equation can be expressed as \(t^2-4t-21=0 \)

Step 2: Applying the quadratic formula

Solve for \( t \) by applying the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -4 \) and \( c = -21 \).

Step 3: Solving for t

Plugging the values into the quadratic formula gives us \( t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-21)}}{2(1)} = 7, -3 \)

Step 4: Reverse Substitution

Substitute \( t \) back into the equation \( t = x - 5 \). Therefore the solutions for \( x \) are \( x = t + 5 = 12, 2 \)

Key Concepts

Algebraic Substitution

When solving equations in algebra, we often encounter expressions that look complex at first glance. Algebraic substitution is a powerful tool that helps simplify these expressions, making them easier to work with.

By introducing a new variable, we can replace a cumbersome part of the equation. For example, in the given exercise, we see the expression \( (x-5)^2 \). Substituting \( t = x - 5 \) simplifies the quadratic equation to \( t^2 - 4t - 21 = 0 \), which is much more straightforward to solve. This technique not only cleans up the equation but also helps reveal the underlying structure that might have been obscured.

Benefits of Algebraic Substitution

• Simplification of complex expressions
• Facilitation of the application of various algebraic methods
• Preparation for more advanced algebraic manipulations in calculus and beyond

Always remember to reverse the substitution at the end to find the solution in terms of the original variable.

Solving Quadratic Equations

Quadratic equations form the backbone of algebra and are identified by the highest power of the variable being squared. They have the standard form \( ax^2 + bx + c = 0 \) where \( a \) is the coefficient of \( x^2 \) and cannot be zero, and \( b \) and \( c \) can be any real number, including zero.

To solve quadratic equations, we can use several methods including factoring, completing the square, and the quadratic formula—\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)—which is especially handy when the equation is difficult to factor. The formula provides a systematic way of finding the roots of any quadratic equation by plugging in the values of \( a \) , \( b \) , and \( c \) from the equation.

When to Use the Quadratic Formula

• When factoring is complex or not possible
• For equations with real or complex coefficients
• When precise solutions are required

Quadratic Equation Roots

The solutions to a quadratic equation are known as its roots. They are the points where the parabola, which is the graph of the quadratic function, intersects the x-axis. These roots can be real or complex and can sometimes be equal, leading to a single intercept point known as a double root.

Using the quadratic formula, we find the roots by calculating the discriminant—\( b^2 - 4ac \)—which determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one real double root.
• If negative, there are two complex roots.

In the exercise, by applying the quadratic formula, we found the roots to be 7 and -3. These correspond to the values that \( t \) can take, which upon reversing our earlier substitution, provide us with the original variable \( x \)’s values of 12 and 2—revealing the points where the parabola crosses the x-axis.