Question

Find the real solutions of each equation. Use a calculator to express any solutions rounded to two decimal places. $$ x^{2 / 3}+4 x^{1 / 3}+2=0 $$

Short answer

The real solution, rounded to two decimal places, is approximately \(x \approx -0.21\).

Step-by-step solution

- Substitute Variable

Let’s make a substitution to simplify the equation. Set \[y = x^{1/3}\]. This means \[y^2 = x^{2/3}\]. Thus, the equation \(x^{2/3} + 4x^{1/3} + 2 = 0\) can be rewritten as \[y^2 + 4y + 2 = 0\].

- Solve the Quadratic Equation

Solve the quadratic equation \[y^2 + 4y + 2= 0\] using the quadratic formula. The quadratic formula is \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], for an equation \[ay^2 + by + c = 0\]. Here, \(a = 1\), \(b = 4\), and \(c = 2\). Substitute these values into the formula:\[y = \frac{-4 \pm \sqrt{16 - 8}}{2}\].

- Simplify the Quadratic Solution

Simplify under the square root and solve for \(y\):\[ y = \frac{-4 \pm \sqrt{8}}{2}\].\[ y = \frac{-4 \pm 2\sqrt{2}}{2}\].This further simplifies to:\[ y = -2 \pm \sqrt{2} \].Thus, the solutions for \(y\) are \[y_1 = -2 + \sqrt{2}\] and \[y_2 = -2 - \sqrt{2}\].

- Reverse the Substitution

Reverse the substitution \( y = x^{1/3}\):\[x^{1/3} = y_1\] and \[x^{1/3} = y_2\].Thus:\[x^{1/3} = -2 + \sqrt{2}\] and \[x^{1/3} = -2 - \sqrt{2}\].

- Check Real Solutions

Evaluate whether the solutions make sense in the context of real numbers. \[-2 - \sqrt{2}\] is negative, and you cannot have a real cube root of a negative number unless the original number is negative. However, \[-2 + \sqrt{2}\] is positive. Confirm the calculation using a calculator to verify only the positive real solution:\[ -2 + \sqrt{2} \approx -2 + 1.41 \approx -0.59\].

- Cube the Positive Solution

Now cube the positive solution value to get the real solution for \(x\): \[x = (-0.59)^3 \approx -0.205\].This will serve as the approximate solution rounded to two decimal places.

Key concepts

Substitution Method in Algebra

The substitution method is a powerful tool in algebra. It helps to simplify complex equations by replacing variables with another one. In our exercise, we used a substitution to transform the original equation \(x^{2/3} + 4x^{1/3} + 2 = 0\). Here's a step-by-step overview:

• We set \( y = x^{1/3} \), making \( y^2 = x^{2/3} \).
• By substituting these into the original equation, we get \( y^2 + 4y + 2 = 0 \).

This simplifies the equation from one involving cube roots to a straightforward quadratic problem. Easy, right? Substitution is great when dealing with polynomials and making them easier to manage. Give it a go next time you face a tricky equation!

Quadratic Formula

The quadratic formula is a crucial concept for solving quadratic equations. Any equation of the form \(ay^2 + by + c = 0\) can be solved using this formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our specific problem, \(a = 1\), \(b = 4\), and \(c = 2\). These are plugged into the formula, giving:

• \( y = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} == \frac{-4 \pm 2\sqrt{2}}{2} \)

This simplifies to \( y = -2 \pm \sqrt{2} \). So, the solutions to the quadratic equation are:

• \( y_1 = -2 + \sqrt{2} \)
• \( y_2 = -2 - \sqrt{2} \)

The quadratic formula is especially useful because it gives you both roots of the quadratic equation directly.

Solving Cubic Equations

Once we have our solutions from the quadratic equation, we need to reverse the substitution \( y = x^{1/3} \) back to find the solutions for \(x\):

• \( x^{1/3} = -2 + \sqrt{2} \)
• \( x^{1/3} = -2 - \sqrt{2} \)

Next, we must determine if these solutions are real. For a cube root to be real, the value inside must be non-negative if we are considering only real numbers. Here, \( -2 - \sqrt{2} \) is negative, making it invalid for cube roots unless we consider complex numbers.

• \( -2 + \sqrt{2} \), however, is positive. Calculating this gives us approximately \( -0.59 \).

Finally, cube this value to find the solution for \( x \). Cubing \( -0.59 \) gives roughly \( -0.205 \), which is the real solution to the original equation. This process shows how cubic equations involve steps of simplification and careful checking of valid solutions.