Problem 136 - College Algebra Essentials

Question

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\text { Solve for } x: x^{\frac{5}{6}}+x^{\frac{2}{3}}-2 x^{\frac{1}{2}}=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 0\) and \(x = 1\).

Step 1: Simplify the Equation

Firstly, substitute \(y = x^{1/6}\) to simplify the equation into a quadratic form. Then the equation becomes : \(y^5 + y^3 - 2y = 0\)

Step 2: Factorize the Equation

Factorize the quadratic equation by grouping common terms together. We have \(y * (y^4 + y^2 - 2) = 0\). This further simplifies to \(y * (y^2 + 2) * (y^2 - 1) = 0\)

Step 3: Solve for Roots

Set each factor equal to zero and solve for 'y'. This gives us \(y^2 + 2 = 0\), \(y^2 - 1 = 0\) and \(y = 0\). From the first equation we get \(y = \pm \sqrt{-2}\), which isn't a real number, so we ignore these roots. From the second equation we get \(y = \pm 1\), and the third equation gives \(y = 0\)

Step 4: Back-Substitute and Solve for 'x'

Now back-substitute \(y = x^{1/6}\) to find the corresponding 'x' values. So the roots of the original equation are \(x = 1\) and \(x = 0\) (remember \(y = 1 => x = 1^6 = 1\) and \(y = 0 => x = 0^6 = 0\) and \(y = -1 => x = (-1)^6 = 1\)

Key Concepts

Simplifying Equations

Simplifying equations involves rewriting them in a form that is easier to handle. In our original equation\[x^{\frac{5}{6}}+x^{\frac{2}{3}}-2 x^{\frac{1}{2}}=0,\]we notice that the exponents make solving the equation directly quite complex.

To simplify, we substitute \(y = x^{1/6}\). Now, every exponent is re-expressed in terms of \(y\):

• \(x^{\frac{5}{6}} = y^5\)
• \(x^{\frac{2}{3}} = y^4\)
• \(x^{\frac{1}{2}} = y^3\)

The equation then becomes \(y^5 + y^3 - 2y = 0\).
This step simplifies the equation by converting it to polynomial form, making it easier to solve.

Factorization

Factorization is a process of breaking down an expression into a product of simpler factors. For the equation \(y^5 + y^3 - 2y = 0\), we can factor by grouping terms that share common factors.

By extracting \(y\) as a common factor, the equation becomes:

• \(y \times (y^4 + y^2 - 2) = 0\)

We further factorize the expression \(y^4 + y^2 - 2\) by recognizing it as a difference of squares:

• \((y^2 + 2)(y^2 - 1)\)

This means the expression can be written as \(y(y^2 + 2)(y^2 - 1) = 0\).
This factorization reveals the roots of the equation and simplifies our problem significantly.

Quadratic Equations

A quadratic equation is a polynomial equation of degree two. However, in our example after simplification and factorization, we dealt with a higher degree equation that resembles multiple quadratic equations:

• \(y^5\)
• \(y^3\)
• \(y^2\)

By factorizing \((y^2 + 2)\) and \((y^2 - 1)\), we treat each as quadratic-like expressions.

Quadratic equations can often be solved using techniques like factoring, completing the square, or using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our scenario, the equation simplifies in such a way that finding roots directly by setting the factors to zero makes the process much easier.

Solving for Roots

Solving for roots means finding the values that make an equation true. For the expression \(y(y^2 + 2)(y^2 - 1) = 0\), set each factor equal to zero and solve:

• \(y = 0\)
• \(y^2 - 1 = 0 \Rightarrow y = \pm 1 \)
• \(y^2 + 2 = 0\) - This leads to non-real numbers so we discard these roots.

Once we have the roots for \(y\), we substitute back: \(y = x^{1/6}\).

• When \(y = 0\), then \(x = 0\).
• When \(y = 1\), then \(x = 1\).
• When \(y = -1\), then \(x = (-1)^6 = 1\).

Understanding this allows us to determine that the solutions to the original equation are \(x = 0\) and \(x = 1\).