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Magazine Chapter 9: Problem 190
Solve. $$15 x^{-2}-26 x^{-1}+8=0$$
Short Answer
Expert verified
The solutions are \( x = \frac{3}{4} \) and \( x = \frac{5}{2} \).
Step by step solution
01
Substitute Variables
First, introduce a substitution to simplify the equation. Let \( y = x^{-1} \). Then, \( x^{-2} = y^2 \). This transforms the equation into a quadratic form: \[ 15 y^2 - 26 y + 8 = 0. \]
02
Identify Coefficients
Identify the coefficients \(a, b,\) and \(c\) in the quadratic equation \(a y^2 + b y + c = 0\). Here, \( a = 15 \), \( b = -26 \), and \( c = 8 \).
03
Apply the Quadratic Formula
Use the quadratic formula to solve for \( y \): \[ y = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}. \] Substituting the values, \[ y = \frac{26 \pm \sqrt{(-26)^2 - 4(15)(8)}}{2(15)}. \]
04
Solve for y
Simplify under the square root: \[ y = \frac{26 \pm \sqrt{676 - 480}}{30} = \frac{26 \pm \sqrt{196}}{30} = \frac{26 \pm 14}{30}. \] Therefore, \( y = \frac{40}{30} = \frac{4}{3} \) or \( y = \frac{12}{30} = \frac{2}{5} \).
05
Substitute Back for x
Recall that \( y = x^{-1} \), so \( x = y^{-1} \). Substitute the found values: For \( y = \frac{4}{3} \), \( x = \frac{3}{4} \). For \( y = \frac{2}{5} \), \( x = \frac{5}{2} \).
06
Final Solution
The solutions to the equation \( 15 x^{-2} - 26 x^{-1} + 8 = 0 \) are \( x = \frac{3}{4} \) and \( x = \frac{5}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
substitution method
The substitution method is a powerful tool to simplify complex equations. In this problem, we start by introducing a substitution: let \( y = x^{-1} \). This changes the difficult equation into a more familiar form. Here鈥檚 why it helps:
Applying this substitution simplifies our equation to \( 15 y^2 - 26 y + 8 = 0 \). This form is more approachable and sets up the stage for the next steps using the quadratic formula.
- It converts a rational equation into a quadratic equation.
- Makes it easier to identify the type of equation we are dealing with.
Applying this substitution simplifies our equation to \( 15 y^2 - 26 y + 8 = 0 \). This form is more approachable and sets up the stage for the next steps using the quadratic formula.
quadratic formula
The quadratic formula is a universal way to solve quadratic equations of the form \( a y^2 + b y + c = 0 \). First, identify the coefficients from our earlier substitution. In this case:
\[ y = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} \].
Substitute \( a, b, \) and \( c \) into the formula:
\[ y = \frac{26 \pm \sqrt{676 - 480}}{30} = \frac{26 \pm \sqrt{196}}{30} = \frac{26 \pm 14}{30} \].
Simplifying further gives us two solutions:
- \( a = 15 \)
- \( b = -26 \)
- \( c = 8 \)
\[ y = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} \].
Substitute \( a, b, \) and \( c \) into the formula:
\[ y = \frac{26 \pm \sqrt{676 - 480}}{30} = \frac{26 \pm \sqrt{196}}{30} = \frac{26 \pm 14}{30} \].
Simplifying further gives us two solutions:
- \( y = \frac{40}{30} = \frac{4}{3} \)
- \( y = \frac{12}{30} = \frac{2}{5} \)
solving rational equations
Solving rational equations involves dealing with equations that include fractions or terms with variable denominators. Initially, our equation \( 15 x^{-2}-26 x^{-1}+8=0 \) is a rational equation. By using a substitution, we transform it into a quadratic equation. This simplifies the solving process:
- Reducing complex fractions to simpler terms.
- Converting the equation to an easier-to-handle quadratic form.
inverse operations
Inverse operations help us 'undo' other operations. In this case, after applying the quadratic formula and finding \( y \) values, we use inverse operations to get back to \( x \). Recall our substitution: \( y = x^{-1} \). So, to find \( x \), we take the reciprocal of \( y \). We had:
- For \( y = \frac{4}{3}, \ x = y^{-1} = \frac{3}{4} \)
- For \( y = \frac{2}{5}, \ x = y^{-1} = \frac{5}{2} \)
