Problem 20 - College Algebra and Calculus: An Applied Approach

Question

Find the real solution(s) of the polynomial equation. Check your solutions. \(x^{6}+3 x^{3}+2=0\)

Step-by-Step Solution

Answer

The real solutions are \( x = -1\) and \( x = \sqrt[3]{-2}\).

Step 1: Substitution

First, replace \(x^{3}\) with a new variable \(y\) so the original polynomial simplifies to \(y^2+3y+2=0\). We have substituted \(y\) for \(x^{3}\) leading to a simpler equation.

Step 2: Solve the quadratic equation

The equation is a quadratic equation that can be solved by the quadratic formula \(y=(-b±\sqrt{b^2-4ac})/(2a)\) after identifying \(a=1\), \(b=3\), and \(c=2\) as the coefficients of the quadratic equation. Solving gives \(y=-1\) and \(y=-2\).

Step 3: Back-substitution

We substituted \(x^{3}\) with \(y\), so \(y\) has to be replaced back with \(x^{3}\) to find the solution for \(x\), hence solving \(x^{3} = -1\) and \(x^{3} = -2\) yields the solutions \(x = -1\), \(x = \sqrt[3]{-2}\).

Step 4: Check the solutions

Substituting each of the solutions back into the original equation confirms that they are indeed solutions.

Key Concepts

Substitution Method

The substitution method involves introducing a new variable to simplify complex equations. In our original equation, we are dealing with a polynomial of the form \(x^6 + 3x^3 + 2 = 0\). Such high-degree equations can be complex to solve directly. To simplify it, we substitute the term \(x^3\) with a new variable, \(y\). This substitution transforms our equation from a sixth-degree polynomial into a more manageable quadratic equation: \(y^2 + 3y + 2 = 0\). By making this clever substitution, we reduce the complexity of the problem, making the next steps of finding solutions more straightforward.

Quadratic Formula

To solve the simplified quadratic equation obtained from the substitution, we employ the quadratic formula. This formula is a powerful tool in algebra that provides solutions to any quadratic equation of the form \(ax^2 + bx + c = 0\) through the expression:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our specific problem, the coefficients are identified as \(a = 1\), \(b = 3\), and \(c = 2\). Substituting these values into the quadratic formula, we solve for \(y\) and obtain the roots \(y = -1\) and \(y = -2\). Using the quadratic formula is a systematic method that ensures accurate solutions for any standard quadratic equation, giving us a concrete basis to proceed further.

Back-substitution

Once we have found the solutions to the simplified quadratic equation, we need to return to our original variable. This process is called back-substitution. We initially set \(y = x^3\), and after solving the quadratic, we found \(y = -1\) and \(y = -2\). Thus, we substitute back to solve for \(x\) with equations \(x^3 = -1\) and \(x^3 = -2\). Solving these gives \(x = -1\) for \(x^3 = -1\), and \(x = \sqrt[3]{-2}\) for \(x^3 = -2\). Back-substitution ensures that the solutions we derived in the simpler context of \(y\) are correctly translated back to the original problem context.

Solution Verification

Verifying solutions is a vital step in solving any equation. It confirms that our solutions are valid for the original equation we set out to solve. After identifying the solutions \(x = -1\) and \(x = \sqrt[3]{-2}\), we need to substitute them back into the original sixth degree equation \(x^6 + 3x^3 + 2 = 0\).

• For \(x = -1\), substituting gives \((-1)^6 + 3(-1)^3 + 2 = 1 - 3 + 2 = 0\).
• For \(x = \sqrt[3]{-2}\), substituting verifies the equation holds true as well.

If both solutions satisfy the original equation, it confirms their correctness, ensuring confidence in the solution process. By verifying our results, we establish that the solutions not only make the simplified quadratic equation true but also satisfy the original polynomial equation.