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Magazine Chapter 14: Problem 19
Solve the given equations algebraically. In Exercise \(10,\) explain your method. $$\left(x^{2}-2 x\right)^{2}-11\left(x^{2}-2 x\right)+24=0$$
Short Answer
Expert verified
The solutions are \( x = 4, -2, 3, -1 \).
Step by step solution
01
Substitute the Intermediate Variable
Let's use substitution to simplify this equation. Let \( y = x^{2} - 2x \). Then the equation becomes \( y^2 - 11y + 24 = 0 \). This substitution helps to reduce the original equation to a simpler quadratic form.
02
Solve the Quadratic Equation
Now, we solve the simplified quadratic equation \( y^2 - 11y + 24 = 0 \). To do this, we can factor the quadratic. We look for two numbers that multiply to 24 and add to -11. These numbers are -8 and -3. Thus, the equation can be factored as \( (y - 8)(y - 3) = 0 \).
03
Find the Values of the Intermediate Variable
Using the factors, set each factor equal to zero: \( y - 8 = 0 \) gives \( y = 8 \), and \( y - 3 = 0 \) gives \( y = 3 \). So the possible values for \( y \) are 8 and 3.
04
Substitute Back to the Original Variable
Recall \( y = x^{2} - 2x \). Substitute \( y = 8 \) back to get \( x^{2} - 2x = 8 \). Rearrange it to form \( x^{2} - 2x - 8 = 0 \).
05
Solve the Quadratic Equation for x (Case y = 8)
Now we solve \( x^{2} - 2x - 8 = 0 \) by factoring. Find two numbers that multiply to -8 and add to -2, which are -4 and 2. Thus, the equation factors to \( (x - 4)(x + 2) = 0 \).
06
Find Values of x (Case y = 8)
Solving \( (x - 4) = 0 \) gives \( x = 4 \) and \( (x + 2) = 0 \) gives \( x = -2 \). Thus, for \( y = 8 \), \( x = 4 \) or \( x = -2 \).
07
Substitute Back to the Original Variable (Case y = 3)
For \( y = 3 \), substitute back to get \( x^{2} - 2x = 3 \). Rearrange this to the quadratic equation \( x^{2} - 2x - 3 = 0 \).
08
Solve the Quadratic Equation for x (Case y = 3)
Factor \( x^{2} - 2x - 3 = 0 \) to \( (x - 3)(x + 1) = 0 \).
09
Find Values of x (Case y = 3)
Solve each factor, \( x - 3 = 0 \) and \( x + 1 = 0 \), to find \( x = 3 \) and \( x = -1 \). So, for \( y = 3 \), \( x = 3 \) or \( x = -1 \).
10
Conclusion
The solutions to the original equation \((x^2-2x)^2-11(x^2-2x)+24=0 \) are \( x = 4, -2, 3, -1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Quadratic Equations
Quadratic equations are a key concept in algebra, generally appearing in the form \( ax^2 + bx + c = 0 \). These equations are called 'quadratic' because the highest exponent of the variable \(x\) is 2. This means that when you graph a quadratic equation, it forms a parabola, which can open upwards or downwards, depending on the coefficients.
Solving quadratic equations can be done using various methods, such as
The solution of a quadratic equation can give you valuable insights about the problem, such as the points where a function crosses the x-axis on a graph. It's key to understanding and solving many algebraic problems.
Solving quadratic equations can be done using various methods, such as
- Factoring
- Completing the square
- Using the quadratic formula
The solution of a quadratic equation can give you valuable insights about the problem, such as the points where a function crosses the x-axis on a graph. It's key to understanding and solving many algebraic problems.
The Method of Factoring
Factoring is a method used to solve quadratic equations by expressing them as a product of simpler expressions. For example, to solve the quadratic equation \( y^2 - 11y + 24 = 0 \), you look for two numbers that multiply to 24 and add up to -11. These numbers are -8 and -3.
By using these numbers, you can factor the quadratic equation to \((y - 8)(y - 3) = 0\). This method simplifies the equation, making it easier to find solutions for y.
Once the equation is factored, you set each factor equal to zero:
Solving each of these gives you the possible values of y, which are 8 and 3 in this example. Factoring is a powerful technique because it turns a complex problem into simpler ones.
By using these numbers, you can factor the quadratic equation to \((y - 8)(y - 3) = 0\). This method simplifies the equation, making it easier to find solutions for y.
Once the equation is factored, you set each factor equal to zero:
- \(y - 8 = 0\)
- \(y - 3 = 0\)
Solving each of these gives you the possible values of y, which are 8 and 3 in this example. Factoring is a powerful technique because it turns a complex problem into simpler ones.
Using Intermediate Variable Substitution
Intermediate variable substitution is a method used to simplify complex algebraic equations by introducing a new variable. This technique is particularly useful when dealing with nested expressions.
In the original exercise, the substitution helps convert the nested expression \((x^2 - 2x)^2 - 11(x^2 - 2x) + 24 = 0\) into a simpler quadratic form by letting \(y = x^2 - 2x\). This results in a new equation \(y^2 - 11y + 24 = 0\) that is easier to solve.
Once you solve for \(y\), you then substitute back the original expression \(y = x^2 - 2x\) to find the values of \(x\). This method breaks down the process and makes a complicated equation manageable.
In the original exercise, the substitution helps convert the nested expression \((x^2 - 2x)^2 - 11(x^2 - 2x) + 24 = 0\) into a simpler quadratic form by letting \(y = x^2 - 2x\). This results in a new equation \(y^2 - 11y + 24 = 0\) that is easier to solve.
Once you solve for \(y\), you then substitute back the original expression \(y = x^2 - 2x\) to find the values of \(x\). This method breaks down the process and makes a complicated equation manageable.
