Problem 115 - Algebra and Trigonometry

Question

Solve: \(2 x^{\frac{2}{3}}-3 x^{\frac{1}{3}}-20=0\)

Step-by-Step Solution

Answer

The solution for the equation \(2 x^{\frac{2}{3}}-3 x^{\frac{1}{3}}-20=0\) is \(x = 64\)

Step 1: Substitute variables

Substitute \(y = x^\frac{1}{3}\) to convert the original equation into the standard quadratic form. Thus, the equation becomes: \(2y^2 - 3y - 20 = 0\)

Step 2: Solve the quadratic equation

Solving this quadratic equation, i.e., \(2y^2 - 3y - 20 = 0\), we use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = -3\), \(c = -20\). Hence, the roots are: \(y = \frac{3 + \sqrt{(#b^2 - #4ac)}}{4}\) and \(y = \frac{3 - \sqrt{(#b^2 - #4ac)}}{4}\). Which simplifies to: \(y = 4\) and \(y = -2.5\).

Step 3: Undo the substitution

Replace \(y\) with \(x^\frac{1}{3}\) in the solutions. Hence, the solutions in \(x\) are: \(x = 4^3 = 64\) and if we substitute \(y = -2.5\) in \(x = y^3\), the equation will not hold true as it will give us a complex number. Hence, the only solution is \(x = 64\).

Key Concepts

Quadratic Equations

Quadratic equations are a special type of algebraic equation where the highest power of the unknown variable is 2. These equations take the general form of:

• \( ax^2 + bx + c = 0 \)

where:

• \( a \), \( b \), and \( c \) are constants
• \( a eq 0 \)
• \( x \) is the variable

Quadratic equations are fundamental in algebra because they model various real-world phenomena such as projectile motion and area calculations. When solving these equations, we can find up to two solutions, also known as roots or zeros, which indicate where the curve of the quadratic function intersects the x-axis. A quadratic equation can have:

• Two distinct real roots
• One real double root
• No real roots (complex roots)

Identifying the type of roots is integral to understanding and solving any quadratic equation, making quadratic equations an essential component of higher-level algebra.

Substitution Method

The substitution method is a valuable strategy in algebra for simplifying complex equations by introducing a new variable. This technique can be especially useful in transforming non-standard equations into a more general form, like converting a transformed quadratic equation to its standard form. In applying substitution, we identify parts of the equation that correspond to a simpler expression, then replace these components with a new variable. This generally simplifies the problem.
In the exercise provided, due to the presence of fractional exponents, we first defined \( y = x^{\frac{1}{3}} \). By substituting \( y \) into the equation \( 2x^{\frac{2}{3}} - 3x^{\frac{1}{3}} - 20 = 0 \), it transforms into a classical quadratic structure:

• \( 2y^2 - 3y - 20 = 0 \)

With this transformation, the once complex expression involving \( x \) and fractional exponents became more manageable, allowing for the equation to be solved using familiar methods like the quadratic formula.

Quadratic Formula

The quadratic formula is a timeless and reliable method for solving quadratic equations. Whenever an equation exists in the form \( ax^2 + bx + c = 0 \), the quadratic formula can determine the roots, or solutions, of the equation:

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

Here's what each term represents:

• \( a \), \( b \), and \( c \) are coefficients from the quadratic equation
• \( \pm \) indicates that there are usually two solutions
• \( \sqrt{b^2 - 4ac} \) is called the discriminant

The discriminant helps to determine the nature of the roots:

• If \( b^2 - 4ac > 0 \), two distinct real roots exist
• If \( b^2 - 4ac = 0 \), there is exactly one real root
• If \( b^2 - 4ac < 0 \), the equation has complex roots

In our provided solution, applying the quadratic formula to \( 2y^2 - 3y - 20 = 0 \) gives us the roots \( y = 4 \) and \( y = -2.5 \). Knowing how the quadratic formula works empowers us to solve even the trickiest quadratic equations confidently.