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Magazine Chapter 11: Problem 29
Solve. \(a^{4}-5 a^{2}+6=0\)
Short Answer
Expert verified
The solutions are \(a = \pm\sqrt{2}, \pm\sqrt{3}\).
Step by step solution
01
Identify the substitution
Observe that the equation \(a^4 - 5a^2 + 6 = 0\) resembles a quadratic equation. We can make this equation easier to solve by letting \(x = a^2\). This changes the equation to \(x^2 - 5x + 6 = 0\).
02
Factor the quadratic equation
The equation \(x^2 - 5x + 6 = 0\) can be factored into \((x - 2)(x - 3) = 0\). This gives us the solutions \(x = 2\) and \(x = 3\).
03
Substitute back to variable 'a'
Since we let \(x = a^2\), substitute back to find \(a\). For \(x = 2\), we have \(a^2 = 2\), giving \(a = \pm\sqrt{2}\). For \(x = 3\), we have \(a^2 = 3\), giving \(a = \pm\sqrt{3}\).
04
List all possible solutions
Combining the results from Step 3, the solutions for \(a\) are \(a = \sqrt{2}, -\sqrt{2}, \sqrt{3}, -\sqrt{3}\).
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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 clever technique used when dealing with polynomial equations that are not immediately quadratic in form but can be made to behave as such. In polynomial equations like \(a^4 - 5a^2 + 6 = 0 \), a "hidden" quadratic can be revealed by using substitution. To do this:
- Identify the part of the equation resembling a quadratic. In this case, notice how \(a^4\) can become \((a^2)^2\), which suggests using a substitution.
- Let \(x = a^2\). This replaces \(a^4\) with \(x^2\) and \(a^2\) itself with \(x\), transforming the original equation into a standard quadratic form: \(x^2 - 5x + 6 = 0\).
Factoring Quadratic Equations
Once the polynomial is simplified into a quadratic form, like \(x^2 - 5x + 6 = 0\), you'll likely need to factor it to find its solutions. Factoring involves breaking down the quadratic into two binomials multiplied together. Here's how you can do it:
- Identify two numbers that multiply to the constant term (6) and add to the middle term coefficient (-5). This would be -2 and -3.
- Rewrite the quadratic as \((x - 2)(x - 3) = 0\).
Square Roots
Now, returning to the original variable \(a\), we solve for \(a\) by taking the square root. From the solved equations \(x = 2\) and \(x = 3\), recall that \(x\) was substituted for \(a^2\). Hence:
- For \(x = 2\), \(a^2 = 2\). Taking the square root gives \(a = \pm \sqrt{2}\).
- For \(x = 3\), \(a^2 = 3\). Similarly, \(a = \pm \sqrt{3}\).
- Remember to consider both positive and negative roots, noted by \(\pm\).
- The square root operation undoes the squaring of \(a^2\), crucial for reverting substitutions.
Algebraic Solutions
By breaking down a polynomial into simpler, solvable parts with algebraic substitutions and operations, we arrive at the algebraic solutions. Here's the complete set of solutions for the original polynomial \(a^4 - 5a^2 + 6 = 0 \):
- From \(x = 2\), solutions \(a = \sqrt{2}\) and \(a = -\sqrt{2}\) arise.
- From \(x = 3\), solutions \(a = \sqrt{3}\) and \(a = -\sqrt{3}\) occur.
- Understanding these methods enables one to solve similar polynomial equations efficiently.
- Each solution step 鈥� substitution, factoring, and square roots 鈥� is fundamental and interlinked.
