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Magazine Chapter 10: Problem 36
Solve each equation. See Example 5 . $$ (a-5)^{2}-4(a-5)-21=0 $$
Short Answer
Expert verified
The solutions are \(a = 12\) and \(a = 2\).
Step by step solution
01
Identify the Substitution
Notice that the expression contains the repeated term \((a-5)\). Let's make a substitution to simplify the equation. Let \( x = a - 5 \). This substitution will make the given equation more manageable.
02
Substitute into the Equation
Replace \((a-5)\) with \(x\) in the equation: \((a-5)^2 - 4(a-5) - 21 = 0\) becomes \(x^2 - 4x - 21 = 0\). This is a quadratic equation in terms of \(x\).
03
Solve the Quadratic Equation
To solve \(x^2 - 4x - 21 = 0\), we can factor the quadratic. Look for two numbers that multiply to -21 and add to -4. The numbers -7 and 3 work: \(x^2 - 4x - 21 = (x - 7)(x + 3) = 0\).
04
Find the Values of x
Set each factor equal to zero and solve for \(x\): - \(x - 7 = 0 \Rightarrow x = 7\) - \(x + 3 = 0 \Rightarrow x = -3\) These are the possible solutions for \(x\).
05
Substitute Back for a
Now substitute back \(x = a - 5\) to find \(a\): - For \(x = 7\), \(a - 5 = 7 \Rightarrow a = 12\) - For \(x = -3\), \(a - 5 = -3 \Rightarrow a = 2\).
06
Verify the Solutions
Plug \(a = 12\) and \(a = 2\) back into the original equation to verify they satisfy the equation:- For \(a = 12\): \((12-5)^2 - 4(12-5) - 21 = (7)^2 - 28 - 21 = 49 - 28 - 21 = 0\).- For \(a = 2\): \((2-5)^2 - 4(2-5) - 21 = (-3)^2 + 12 - 21 = 9 + 12 - 21 = 0\).Both values satisfy the equation, confirming they are correct solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a fundamental concept in algebra. It is an equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). These equations are recognized by the squared term, \( x^2 \). Quadratic equations can appear in diverse forms but can always be solved through specific methods. A quadratic equation can represent a parabola when graphed.
The equation from the exercise, \((a-5)^2 - 4(a-5) - 21 = 0\), is indeed a quadratic equation after substitution, with \( x = a - 5 \). This substitution transforms the problem into a much cleaner form, making it easier to solve. The goal usually is to find the values of \( x \) (in this case \( a \)) that satisfy the equation.
The equation from the exercise, \((a-5)^2 - 4(a-5) - 21 = 0\), is indeed a quadratic equation after substitution, with \( x = a - 5 \). This substitution transforms the problem into a much cleaner form, making it easier to solve. The goal usually is to find the values of \( x \) (in this case \( a \)) that satisfy the equation.
Factoring Quadratics
Factoring is one of the most straightforward ways to solve a quadratic equation when it's set to zero. In our problem, the equation after substitution becomes \( x^2 - 4x - 21 = 0 \). Factoring transforms the quadratic into a product of two binomials: \((x - 7)(x + 3) = 0\).
To factor, you need two numbers that multiply to the constant term, -21, and add to the coefficient of the middle term, -4. Here, -7 and 3 are the numbers that work:
To factor, you need two numbers that multiply to the constant term, -21, and add to the coefficient of the middle term, -4. Here, -7 and 3 are the numbers that work:
- -7 + 3 = -4 (the coefficient of \( x \))
- -7 \( \times \) 3 = -21 (the constant term)
Algebraic Substitution
Algebraic substitution is a technique used to simplify complex equations. It involves replacing a repeated term or expression with a new variable. In this exercise, the substitution \( x = a - 5 \) helps manage the equation by reducing its complexity.
This is particularly useful when an expression appears repeatedly, as it can transform the equation into a more solvable form. By letting \( x = a - 5 \), the original equation \((a-5)^2 - 4(a-5) - 21 = 0\) simplifies to \( x^2 - 4x - 21 = 0 \).
Once the quadratic is solved, you substitute back to find the original variable, \( a \). Each solution for \( x \) indicates a possible value for \( a \). For this problem, \( x = 7 \) leads to \( a = 12 \), and \( x = -3 \) results in \( a = 2 \). Substitution not only simplifies the calculation but also highlights the beauty and utility of algebraic problem-solving techniques.
This is particularly useful when an expression appears repeatedly, as it can transform the equation into a more solvable form. By letting \( x = a - 5 \), the original equation \((a-5)^2 - 4(a-5) - 21 = 0\) simplifies to \( x^2 - 4x - 21 = 0 \).
Once the quadratic is solved, you substitute back to find the original variable, \( a \). Each solution for \( x \) indicates a possible value for \( a \). For this problem, \( x = 7 \) leads to \( a = 12 \), and \( x = -3 \) results in \( a = 2 \). Substitution not only simplifies the calculation but also highlights the beauty and utility of algebraic problem-solving techniques.
