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Magazine Chapter 3: Problem 38
Find all real and imaginary solutions to each equation. Check your answers. $$x^{4}-x^{2}-12=0$$
Short Answer
Expert verified
Real solutions: \(x = 2, -2\). Imaginary solutions: \(x = i \sqrt{3}, -i \sqrt{3}\).
Step by step solution
01
Substitute {\(y\) for \(x^2\)}
Let \(y = x^2\). This transforms the original equation \(x^4 - x^2 - 12 = 0\) into a quadratic equation: \(y^2 - y - 12 = 0\).
02
Solve the quadratic equation
Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, \(a = 1\), \(b = -1\), and \(c = -12\). Plug these values into the formula to get the solutions for \(y\).
03
Determine the discriminant
Calculate the discriminant \(b^2 - 4ac\). With \(a = 1\), \(b = -1\), and \(c = -12\), the discriminant is \((-1)^2 - 4(1)(-12) = 1 + 48 = 49\).
04
Calculate the roots of the quadratic equation
The roots are given by \(\frac{1 \pm \sqrt{49}}{2} = \frac{1 \pm 7}{2}\). This yields two solutions: \(y = 4\) and \(y = -3\).
05
Substitute back \(x^2\) for \(y\)
Remember that \(y = x^2\). So, we have \(x^2 = 4\) and \(x^2 = -3\).
06
Solve for \(x\) where \(x^2 = 4\)
Take the square root of both sides. \(x = \pm 2\). So the real solutions are \(x = 2\) and \(x = -2\).
07
Solve for \(x\) where \(x^2 = -3\)
Take the square root of both sides. \(x = \pm i \sqrt{3} \). So the imaginary solutions are \(x = i \sqrt{3}\) and \(x = -i \sqrt{3}\).
08
Check the solutions
Substitute each solution back into the original equation to verify. \(x = 2\), \(x = -2\), \(x = i \sqrt{3}\), and \(x = -i \sqrt{3}\) all satisfy the equation \(x^4 - x^2 - 12 = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic substitution
Quadratic substitution is a technique to simplify solving higher degree polynomial equations. By substituting a simpler variable for a more complex expression, we can transform a complex equation into something easier to manage. In the given exercise, we encounter the polynomial equation 饾懃^4 鈭� 饾懃^2 鈭� 12 = 0. Solving this directly is difficult, but by substituting 饾懄 for 饾懃^2, we transform it into a quadratic equation: 饾懄^2 鈭� 饾懄 鈭� 12 = 0.
This new equation is easier to handle because it's a quadratic, and we can apply familiar methods like factoring or the quadratic formula. The substitution step is crucial because it makes the problem solvable with standard techniques, showing why quadratic substitution is a powerful tool in algebra.
This new equation is easier to handle because it's a quadratic, and we can apply familiar methods like factoring or the quadratic formula. The substitution step is crucial because it makes the problem solvable with standard techniques, showing why quadratic substitution is a powerful tool in algebra.
discriminant in quadratic formula
The discriminant in the quadratic formula helps determine the nature of the roots of a quadratic equation. In our case, the equation 饾懃^4 鈭� 饾懃^2 鈭� 12 = 0 transforms into 饾懄^2 鈭� 饾懄 鈭� 12 = 0 after substitution. To find the roots, we use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, the discriminant (\b^2 - 4ac\b) is crucial. It tells us about the nature of our roots without solving the equation completely. For the quadratic 饾懄^2 鈭� 饾懄 鈭� 12 = 0, we plug in 饾憥 = 1, 饾憦 = 鈭�1, and 饾憪 = 鈭�12. This gives us:
\(-1^2 - 4(1)(-12) = 1 + 48 = 49\)
Since the discriminant is positive (49), it indicates two distinct real roots. This step not only helps in determining root characteristics but also guides us on how to proceed with the solution.
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, the discriminant (\b^2 - 4ac\b) is crucial. It tells us about the nature of our roots without solving the equation completely. For the quadratic 饾懄^2 鈭� 饾懄 鈭� 12 = 0, we plug in 饾憥 = 1, 饾憦 = 鈭�1, and 饾憪 = 鈭�12. This gives us:
\(-1^2 - 4(1)(-12) = 1 + 48 = 49\)
Since the discriminant is positive (49), it indicates two distinct real roots. This step not only helps in determining root characteristics but also guides us on how to proceed with the solution.
real and imaginary solutions
Real and imaginary solutions are fundamental concepts in solving polynomial equations. These solutions signify the different types of roots an equation can have. For our transformed equation 饾懄^2 鈭� 饾懄 鈭� 12 = 0, solving it gives us 饾懄 = 4 and 饾懄 = -3.
When we substitute back 饾懄 for 饾懃^2, we get two equations: 饾懃^2 = 4 and 饾懃^2 = -3.
- For 饾懃^2 = 4, taking the square root of both sides gives us real solutions: 饾懃 = 卤2.
- For 饾懃^2 = -3, the solutions are imaginary. Taking the square root, we get: 饾懃 = 卤i鈭�3.
Imaginary solutions occur when we're dealing with the square root of a negative number. In this case, '饾憱' represents the imaginary unit (鈭�-1).
Understanding the distinction between real and imaginary solutions helps in comprehensively solving polynomial equations and recognizing the broad spectrum of possible solutions.
When we substitute back 饾懄 for 饾懃^2, we get two equations: 饾懃^2 = 4 and 饾懃^2 = -3.
- For 饾懃^2 = 4, taking the square root of both sides gives us real solutions: 饾懃 = 卤2.
- For 饾懃^2 = -3, the solutions are imaginary. Taking the square root, we get: 饾懃 = 卤i鈭�3.
Imaginary solutions occur when we're dealing with the square root of a negative number. In this case, '饾憱' represents the imaginary unit (鈭�-1).
Understanding the distinction between real and imaginary solutions helps in comprehensively solving polynomial equations and recognizing the broad spectrum of possible solutions.
