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Magazine Chapter 10: Problem 59
Solve each equation. $$ \left(y^{2}-9\right)^{2}+2\left(y^{2}-9\right)-99=0 $$
Short Answer
Expert verified
The solutions are \( y = 3\sqrt{2} \) and \( y = -3\sqrt{2} \).
Step by step solution
01
Substitute Variable
Let\( a = y^2 - 9 \). Then the equation\( \left( y^2 - 9 \right)^2 + 2 \left( y^2 - 9 \right) - 99 = 0 \) can be rewritten as:\[ a^2 + 2a - 99 = 0 \] This substitution simplifies solving the original equation into solving a quadratic equation in terms of\( a \).
02
Solve the Quadratic Equation
The quadratic equation\( a^2 + 2a - 99 = 0 \) can be solved using the quadratic formula, which states that for any quadratic equation of the form\( ax^2 + bx + c = 0 \), the solutions are given by:\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case,\( a = 1 \),\( b = 2 \),\( c = -99 \), which gives:\[ a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-99)}}{2 \cdot 1} \] Simplifying this gives:\[ a = \frac{-2 \pm \sqrt{4 + 396}}{2} \] \[ a = \frac{-2 \pm \sqrt{400}}{2} \] \[ a = \frac{-2 \pm 20}{2} \] Thus, the solutions are\( a = 9 \) or\( a = -11 \).
03
Substitute Back to Original Variable
Recall that\( a = y^2 - 9 \). We will substitute back\( a = 9 \) and\( a = -11 \) to solve for\( y \). First solve for\( y^2 - 9 = 9 \):\[ y^2 = 18 \] \[ y = \pm \sqrt{18} \] \[ y = \pm 3\sqrt{2} \] Now, solve for\( y^2 - 9 = -11 \):\[ y^2 = -2 \] Since the square of a real number cannot be negative, this equation has no real solutions. Thus, the only solutions are from\( y^2 = 18 \).
04
Finalize the Solution
Thus, the solution to the equation\( \left(y^{2}-9\right)^{2}+2\left(y^{2}-9\right)-99=0 \) in terms of\( y \) is\( y = 3\sqrt{2} \) and\( y = -3\sqrt{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
To solve complex quadratic equations, the substitution method can be a savior. When facing a daunting equation like \((y^2-9)^2 + 2(y^2-9) - 99 = 0\), direct solving might seem overwhelming. Instead of trying to untangle it as is, we introduce a simpler variable to ease the process. Here, we substitute \(a = y^2 - 9\). This clever trick transforms the original equation into a more manageable form: \(a^2 + 2a - 99 = 0\).Using substitution:
- Identifies a recurrent component in the equation.
- Reduces a complicated expression to a standard quadratic form.
- Makes it easier to apply further techniques like the quadratic formula.
Quadratic Formula
The quadratic formula is a powerful tool for finding solutions to any quadratic equation of the form \(ax^2 + bx + c = 0\). This formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), directly provides the solutions by leveraging the coefficients of the equation.In our exercise:
- The simplified equation is \(a^2 + 2a - 99 = 0\).
- Here, \(a = 1\), \(b = 2\), and \(c = -99\).
- Plugging these values in, the formula gives \(a = \frac{-2 \pm \sqrt{4 + 396}}{2}\), leading to solutions \(a = 9\) and \(a = -11\).
Real Solutions
A fundamental concept in solving equations is understanding what constitutes a real solution. A solution is considered real if it can be expressed without involving imaginary numbers.During our problem-solving:
- After solving \(a^2 + 2a - 99 = 0\), we found \(a = 9\) and \(a = -11\).
- Substituting back, \(y^2 - 9 = 9\) gives \(y = \pm 3\sqrt{2}\), which are real numbers.
- For \(y^2 - 9 = -11\), it results in \(y^2 = -2\), leading to imaginary solutions, thus no real solution here.
Variable Substitution
The power of variable substitution lies in its ability to transform a problem into a more familiar form. In the given exercise, we see how substituting \(a = y^2 - 9\) makes the equation much easier to tackle.Benefits of variable substitution:
- Breaks down complex expressions into simpler, standard parts.
- Facilitates easier manipulation, allowing the use of well-known formulas and techniques.
- After solving, the original variable is revisited, merging simplicity with the original context.
