Problem 87
Question
Solve each equation. $$10 x^{-2}+33 x^{-1}-7=0$$
Step-by-Step Solution
Verified Answer
The solutions are \( x = 5 \) and \( x = -\frac{2}{7} \).
1Step 1: Introduce a substitution to simplify the equation
Let’s set a substitution where we let \(y = x^{-1}\). This changes the equation in terms of \(y\).
2Step 2: Rewrite the equation
Rewrite the equation with the substitution: \[10y^2 + 33y - 7 = 0\]
3Step 3: Solve the quadratic equation
Use the quadratic formula, \( y = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \), where \(a = 10\), \(b = 33\), and \(c = -7\). Calculate the discriminant: \[b^2 - 4ac = 33^2 - 4(10)(-7) = 1089 + 280 = 1369\]So the roots are: \[y = \frac{-33 \, \pm \, \sqrt{1369}}{20}\]Which simplifies to: \[y = \frac{-33 \, \pm \, 37}{20}\]Thus, the solutions are:\[y = \frac{4}{20} = \frac{1}{5}\] and \[y = \frac{-70}{20} = -3.5\]
4Step 4: Back-substitute the values to find x
Recall that \(y = x^{-1}\) so:If \( y = \frac{1}{5} \), then \( x = 5 \).If \( y = -3.5 \), then \( x = -\frac{2}{7} \).
Key Concepts
substitution methodquadratic formulaback-substitution
substitution method
The substitution method is a powerful tool for solving complex equations by transforming them into simpler ones. For the given equation, we are starting with \(10 x^{-2} + 33 x^{-1} - 7 = 0\). Solving this directly can be difficult. So, we use substitution to convert it into an easier form.
We set \(y = x^{-1}\). This means everywhere we see \(x^{-1}\), we replace it with \(y\). Doing this turns our original equation into \(10y^2 + 33y - 7 = 0\).
Notice how more manageable this new equation is. Using substitution can make solving equations less intimidating and easier to understand.
We set \(y = x^{-1}\). This means everywhere we see \(x^{-1}\), we replace it with \(y\). Doing this turns our original equation into \(10y^2 + 33y - 7 = 0\).
Notice how more manageable this new equation is. Using substitution can make solving equations less intimidating and easier to understand.
quadratic formula
The quadratic formula is instrumental in solving quadratic equations, like the one we obtained through substitution: \(10y^2 + 33y - 7 = 0\). The formula to solve such an equation is:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^2 + bx + c = 0\). For our equation: \(a = 10\), \(b = 33\), and \(c = -7\).
We first calculate the discriminant, \(b^2 - 4ac\):
\[ 33^2 - 4(10)(-7) = 1089 + 280 = 1369 \]
With the discriminant calculated, we can now find \(y\):
\[ y = \frac{-33 \pm \sqrt{1369}}{20} \]
This solves to:
\[ y = \frac{-33 + 37}{20} = \frac{4}{20} = \frac{1}{5} \]
\[ y = \frac{-33 - 37}{20} = \frac{-70}{20} = -3.5 \]
Hence, the solutions for \(y\) are \(\frac{1}{5}\) and \(-3.5\). Using the quadratic formula consistently helps find roots of quadratic equations efficiently.
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^2 + bx + c = 0\). For our equation: \(a = 10\), \(b = 33\), and \(c = -7\).
We first calculate the discriminant, \(b^2 - 4ac\):
\[ 33^2 - 4(10)(-7) = 1089 + 280 = 1369 \]
With the discriminant calculated, we can now find \(y\):
\[ y = \frac{-33 \pm \sqrt{1369}}{20} \]
This solves to:
\[ y = \frac{-33 + 37}{20} = \frac{4}{20} = \frac{1}{5} \]
\[ y = \frac{-33 - 37}{20} = \frac{-70}{20} = -3.5 \]
Hence, the solutions for \(y\) are \(\frac{1}{5}\) and \(-3.5\). Using the quadratic formula consistently helps find roots of quadratic equations efficiently.
back-substitution
After solving for \(y\), we must revert to the original variable \(x\) using back-substitution. Remember, our substitution was \(y = x^{-1}\), i.e., \(y = \frac{1}{x}\).
For \(y = \frac{1}{5}\):
We know \(\frac{1}{x} = \frac{1}{5}\), so \(x = 5\).
For \(y = -3.5\):
We know \(\frac{1}{x} = -3.5\), so \(x = \frac{1}{-3.5} = -\frac{2}{7}\).
Thus, the final solutions for \(x\) are \(x = 5\) and \(x = -\frac{2}{7}\).
The back-substitution process transforms back the solutions we obtained for the substituted variable into solutions for the original variable. This ensures our final answers are relevant to the initial problem.
For \(y = \frac{1}{5}\):
We know \(\frac{1}{x} = \frac{1}{5}\), so \(x = 5\).
For \(y = -3.5\):
We know \(\frac{1}{x} = -3.5\), so \(x = \frac{1}{-3.5} = -\frac{2}{7}\).
Thus, the final solutions for \(x\) are \(x = 5\) and \(x = -\frac{2}{7}\).
The back-substitution process transforms back the solutions we obtained for the substituted variable into solutions for the original variable. This ensures our final answers are relevant to the initial problem.
Other exercises in this chapter
Problem 87
Write each statement as an absolute value equation or inequality. \(r\) is no less than 1 unit from 29.
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Solve each rational inequality. Write each solution set in interval notation.4 $$\frac{2 x-3}{x^{2}+1} \geq 0$4
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Evaluate the discriminant for each equation. Then use it to predict the number of distinct solutions, and whether they are rational, irrational, or non real com
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Write each statement as an absolute value equation or inequality. \(q\) is no more than 8 units from 22.
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