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Magazine Chapter 11: Problem 56
Solve each equation. Check the solutions. \(z^{4}+72=17 z^{2}\)
Short Answer
Expert verified
The solutions are \( z = 3, -3, 2\sqrt{2}, -2\sqrt{2} \).
Step by step solution
01
Move all terms to one side
Rewrite the equation by moving all terms to one side such that the equation equals zero: \( z^4 + 72 - 17z^2 = 0 \)
02
Substitute and simplify
Let \( u = z^2 \). Substitute \( u \) into the equation: \( u^2 + 72 - 17u = 0 \). Now, reorder it as: \( u^2 - 17u + 72 = 0 \).
03
Solve the quadratic equation
Solve the quadratic equation \( u^2 - 17u + 72 = 0 \). This can be done using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 1 \), \( b = -17 \), and \( c = 72 \): \[ u = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 1 \cdot 72}}{2 \cdot 1} \] Calculating the discriminant: \( 17^2 - 4 \cdot 72 = 289 - 288 = 1 \), So the solutions are: \[ u = \frac{17 \pm 1}{2} \] This gives \( u = 9 \) and \( u = 8 \).
04
Back-substitute to find \( z \)
Substitute back \( u = z^2 \): For \( u = 9 \): \( z^2 = 9 \) \( z = \pm 3 \)For \( u = 8 \): \( z^2 = 8 \) \( z = \pm \sqrt{8} = \pm 2\sqrt{2} \)
05
Verify the solutions
Plug each solution back into the original equation to verify:For \( z = 3 \): \[ 3^4 + 72 = 17 \cdot 3^2 \Rightarrow 81 + 72 = 153 \Rightarrow 153 = 153 \]This is correct.For \( z = -3 \): \[ (-3)^4 + 72 = 17 \cdot (-3)^2 \Rightarrow 81 + 72 = 153 \Rightarrow 153 = 153 \]This is correct.For \( z = 2\sqrt{2} \): \[ (2\sqrt{2})^4 + 72 = 17 \cdot (2\sqrt{2})^2 \Rightarrow 64 + 72 = 136 \Rightarrow 136 = 136 \]This is correct.For \( z = -2\sqrt{2} \): \[ (-2\sqrt{2})^4 + 72 = 17 \cdot (-2\sqrt{2})^2 \Rightarrow 64 + 72 = 136 \Rightarrow 136 = 136 \]This is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Substitution
When faced with a polynomial equation, sometimes it becomes easier to solve it if you reduce the degree of the polynomial. For example, if you have a quartic equation (degree 4), you can use substitution to turn it into a quadratic equation (degree 2).
In this exercise, we start with the equation: \(z^{4}+72=17z^{2}\).
To simplify, we substitute \(u = z^{2}\). By doing this substitution, our original quartic equation \(z^{4}+72=17z^{2}\) converts into a quadratic equation: \(u^{2} + 72 - 17u = 0\).
Once the equation is quadratic, it becomes manageable using methods we can apply to quadratic equations.
In this exercise, we start with the equation: \(z^{4}+72=17z^{2}\).
To simplify, we substitute \(u = z^{2}\). By doing this substitution, our original quartic equation \(z^{4}+72=17z^{2}\) converts into a quadratic equation: \(u^{2} + 72 - 17u = 0\).
Once the equation is quadratic, it becomes manageable using methods we can apply to quadratic equations.
Polynomial Roots
Finding the roots of a polynomial means solving the equation to find the values where the polynomial equals zero.
Once the equation is simplified into a quadratic form using a substitution like \(u = z^{2}\), we need to solve the quadratic equation to find 'u'.
The quadratic equation in this problem, \(u^{2} - 17u + 72\), can be solved using various methods such as factoring, completing the square, or using the quadratic formula. For this specific problem, the quadratic formula emerges as a straightforward choice.
Once the equation is simplified into a quadratic form using a substitution like \(u = z^{2}\), we need to solve the quadratic equation to find 'u'.
The quadratic equation in this problem, \(u^{2} - 17u + 72\), can be solved using various methods such as factoring, completing the square, or using the quadratic formula. For this specific problem, the quadratic formula emerges as a straightforward choice.
Quadratic Formula
The quadratic formula is an essential tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
For our problem, we apply it to the equation \(u^{2} - 17u + 72 = 0\). Here, \(a = 1\), \(b = -17\), and \(c = 72\).
First, we calculate the discriminant: \(-17)^{2} - 4 \times 1 \times 72 = 289 - 288 = 1\).
Thus, our solutions for 'u' are \(u = \frac{17 \pm 1}{2}\), which gives us \(u = 9\) and \(u = 8\).
These 'u' values need to be back-substituted to find 'z'.
For our problem, we apply it to the equation \(u^{2} - 17u + 72 = 0\). Here, \(a = 1\), \(b = -17\), and \(c = 72\).
First, we calculate the discriminant: \(-17)^{2} - 4 \times 1 \times 72 = 289 - 288 = 1\).
Thus, our solutions for 'u' are \(u = \frac{17 \pm 1}{2}\), which gives us \(u = 9\) and \(u = 8\).
These 'u' values need to be back-substituted to find 'z'.
Verifying Solutions
After finding potential solutions, it's crucial to verify them in the original equation to ensure they are correct.
From our problem, we find \(u = 9\) and \(u = 8\), which translates back to \(z^{2} = 9\) and \(z^{2} = 8\).
Specifically, for \(z^{2} = 9\), we get \(z = \pm3\). For \(z^{2} = 8\), we get \(z = \pm2\sqrt{2}\).
Plugging each solution into the original quartic equation \(z^{4}+72=17z^{2}\) confirms they satisfy the equation:
- For \(z = 3\) and \(z = -3\): \(81 + 72 = 153\) matches \(17 \cdot 9 = 153\).
- For \(z = 2\sqrt{2}\) and \(z = -2\sqrt{2}\): \(64 + 72 = 136\) matches \(17 \cdot 8 = 136\).
Verifying solutions gives confidence that the solutions are correct.
From our problem, we find \(u = 9\) and \(u = 8\), which translates back to \(z^{2} = 9\) and \(z^{2} = 8\).
Specifically, for \(z^{2} = 9\), we get \(z = \pm3\). For \(z^{2} = 8\), we get \(z = \pm2\sqrt{2}\).
Plugging each solution into the original quartic equation \(z^{4}+72=17z^{2}\) confirms they satisfy the equation:
- For \(z = 3\) and \(z = -3\): \(81 + 72 = 153\) matches \(17 \cdot 9 = 153\).
- For \(z = 2\sqrt{2}\) and \(z = -2\sqrt{2}\): \(64 + 72 = 136\) matches \(17 \cdot 8 = 136\).
Verifying solutions gives confidence that the solutions are correct.
