Chapter 7: Problem 115
Solve: \(2 x^{\frac{2}{3}}-3 x^{\frac{1}{3}}-20=0\)
Short Answer
Expert verified
The solution for the equation \(2 x^{\frac{2}{3}}-3 x^{\frac{1}{3}}-20=0\) is \(x = 64\)
Step by step solution
01
Substitute variables
Substitute \(y = x^\frac{1}{3}\) to convert the original equation into the standard quadratic form. Thus, the equation becomes: \(2y^2 - 3y - 20 = 0\)
02
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 \(y = -2.5\).
03
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 \(x = 64\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \( ax^2 + bx + c = 0 \)
- \( a \), \( b \), and \( c \) are constants
- \( a eq 0 \)
- \( x \) is the variable
- Two distinct real roots
- One real double root
- No real roots (complex roots)
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:
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:
- \( 2y^2 - 3y - 20 = 0 \)
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:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( a \), \( b \), and \( c \) are coefficients from the quadratic equation
- \( \pm \) indicates that there are usually two solutions
- \( \sqrt{b^2 - 4ac} \) is called the discriminant
- 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