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Chapter 8: Problem 36

Solve. $$t^{1 / 2}+3 t^{1 / 4}+2=0$$

Short Answer

Expert verified
No real solutions.

Step by step solution

01

Substitution

Let us substitute the variable to simplify the equation. Set u = t^{1/4}. Therefore, the equation becomes u^2 + 3u + 2 = 0.
02

Solve the quadratic equation

Now solve the quadratic equation u^2 + 3u + 2 = 0. Factoring, we get (u + 1)(u + 2) = 0. Therefore, u = -1 or u = -2.
03

Substitute back

Substitute back the value of u so t^{1/4} = -1 or t^{1/4} = -2.
04

Address possible real solutions

Since t^{1/4} cannot be negative for a real number t, there are no real solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

substitution method
The substitution method can make solving complex equations easier by changing variables. In this exercise, we started with the equation \( t^{1/2} + 3t^{1/4} + 2 = 0 \). This looks challenging.
To simplify, we set a new variable, \( u \), where \( u = t^{1/4} \). This changes our equation to \( u^2 + 3u + 2 = 0 \).
This new equation is a standard quadratic equation. By substituting, we turned a complex equation into a simpler one.
factoring
Factoring is a method used to solve quadratic equations by breaking them down into simpler expressions.
Once we substituted and got the equation \( u^2 + 3u + 2 = 0 \), the next step was to factor it. Factoring involves finding two binomials that multiply to give the quadratic equation.
For \( u^2 + 3u + 2 \), we found that \( (u + 1)(u + 2) = 0 \) was the factored form.
This gave us the solutions \( u = -1 \) and \( u = -2 \).
real numbers
Real numbers are any numbers that are not imaginary; they include all rational and irrational numbers.
When solving equations, it's important to check if the solutions fall within the set of real numbers.
In our exercise, we found that \( u = -1 \) and \( u = -2 \).
Then, we substituted back to get \( t^{1/4} = -1 \) and \( t^{1/4} = -2 \).
Since negative numbers can't result from raising a real number to the fourth power, these solutions are not real. Therefore, they aren't valid solutions in the context of real numbers.
no real solutions
After finding the solutions in the factored form and substituting back, we reached \( t^{1/4} = -1 \) and \( t^{1/4} = -2 \).
Both are not possible for any real number \( t \). Because raising a real number to the fourth power can never yield a negative result, these values cannot be true.
Thus, we concluded there are no real solutions for the original equation \( t^{1/2} + 3t^{1/4} + 2 = 0 \). This is an important finding, emphasizing the necessity to verify if solutions are within the scope of real numbers.

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