Problem 39 - College Algebra

Question

Solve each equation with rational exponents. Check all proposed solutions. $$\left(x^{2}-x-4\right)^{\frac{3}{4}}-2-6$$

Step-by-Step Solution

Answer

The solution to the equation \(\left(x^{2}-x-4\right)^{\frac{3}{4}}-2-6=0\) is \(x=5\).

Step 1: Simplify the equation

First, simplify the equation by combining like terms. The equation \(\left(x^{2}-x-4\right)^{\frac{3}{4}}-2-6\) can be simplified to \(\left(x^{2}-x-4\right)^{\frac{3}{4}}-8=0\)

Step 2: Isolate the exponential term

In order to solve for \(x\), the first aim is to isolate the term with the exponent. By adding 8 to each side of the equation: \(\left(x^{2}-x-4\right)^{\frac{3}{4}}=8\)

Step 3: Remove the rational exponent

To remove the fractional exponent, raise both sides of the equation to the 4/3 power: \((\left(x^{2}-x-4\right)^{\frac{3}{4}})^{\frac{4}{3}}=8^{\frac{4}{3}},\) which simplifies to \(x^{2}-x-4=8^{\frac{4}{3}}\).

Step 4: Solve for x

Solve the resulting quadratic equation \(x^{2}-x-4=8^{\frac{4}{3}}\). First, calculate \(8^{\frac{4}{3}}\) to get a numerical value. This results in the quadratic equation \(x^{2}-x-4=16\). By rearranging the terms, we get \(x^{2}-x-20=0\). Factorising this equation, we get \((x-5)(x+4)=0\). Therefore, \(x=5\) or \(x=-4\).

Step 5: Verify solutions

Substitute the obtained solutions back into the original question to ensure they are valid. It turns out that \(x=5\) is the only valid solution, since \(x=-4\) results in a negative number under the radical sign in the original equation.

Key Concepts

Rational Exponents

Understanding rational exponents is critical when solving equations that involve fractional powers. Exponents of this form are written as \(a^{m/n}\), where \(a\) is the base, \(m\) is the numerator, and \(n\) is the denominator of the exponent. The expression represents the \(n\)-th root of \(a\) raised to the \(m\)-th power. This can also be written as \(\sqrt[n]{a^m}\) or \(\left(\sqrt[n]{a}\right)^m\).

In the provided exercise, we have the term \(\left(x^{2}-x-4\right)^{\frac{3}{4}}\). To comprehend this, think of it as taking the fourth root of \(x^{2}-x-4\) and then cubing the result. This concept is essential because it forms the basis of removing the exponent to solve the equation. When solving, we perform the inverse operation, which means raising both sides of the equation to the reciprocal of the fractional exponent, effectively canceling it out.

An awareness of this order of operations is essential. Neglecting to correctly manipulate the rational exponent can result in a misinterpretation of the problem and a wrong solution.

Isolate Exponential Term

Isolating the exponential term in an equation allows us to solve for the variable that's inside the exponent. This step is all about achieving a clearer view by ensuring that the term with a variable exponent stands alone on one side of the equation. In our exercise, the term with the exponent is \(\left(x^{2}-x-4\right)^{\frac{3}{4}}\).

To isolate this term, other elements on the same side of the equation need to be moved across the equals sign. This is typically achieved through addition or subtraction. In this case, we add 8 to both sides to cancel the -8 on the left, leading to \(\left(x^{2}-x-4\right)^{\frac{3}{4}}=8\). The process of isolation is critical; without it, it would not be possible to correctly apply the inverse operations required to remove the rational exponent and solve for \(x\).

It's vital for students to grasp this concept, as isolating terms is a common technique not only in solving equations with exponents but also in various types of algebraic manipulations.

Quadratic Equations

Quadratic equations are fundamental in algebra and have the general form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are constants. These equations can be solved by factoring, completing the square, or using the quadratic formula. In the given problem, once we have removed the rational exponent and simplified, we end up with a quadratic equation \(x^{2}-x-20=0\).

Here, factoring is the method used: breaking down the quadratic into two binomials that multiply to give the original quadratic. For \(x^{2}-x-20\), we find factors that multiply to -20 and add to -1, which are -5 and 4. Therefore, we get \(x-5)(x+4)=0\), leading to the solutions \(x=5\) and \(x=-4\).

However, it's important to remember that not all solutions derived algebraically will fit the original equation context. Checking each solution in the initial equation is essential to confirm validity. In the exercise, we check and see that only \(x=5\) is a valid solution.