/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Solve. $$ \left(x^{2}-5 x-1\... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 68

Solve. $$ \left(x^{2}-5 x-1\right)^{2}-18\left(x^{2}-5 x-1\right)+65=0 $$

Short Answer

Expert verified
The solutions are \( x = 7 \), \( x = -2 \), \( x = 6 \), and \( x = -1 \).

Step by step solution

01

- Substitute a New Variable

Let\( y = x^{2}-5x-1 \). This simplifies the equation to: \[ y^2 - 18y + 65 = 0 \]
02

- Use the Quadratic Formula

The quadratic formula is: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a=1\), \(b=-18\), and \(c=65\). Substitute these values into the formula: \[ y = \frac{-(-18) \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 65}}{2 \cdot 1} \]This simplifies to: \[ y = \frac{18 \pm \sqrt{324 - 260}}{2} \] \[ y = \frac{18 \pm \sqrt{64}}{2} \] \[ y = \frac{18 \pm 8}{2} \]
03

- Solve for y

Calculate the two values for \(y\): \[ y = \frac{18 + 8}{2} = 13 \] \[ y = \frac{18 - 8}{2} = 5 \]Thus, \( y = 13 \) or \( y = 5 \).
04

- Back-substitute the Original Variable

Recall \( y = x^2 - 5x - 1 \). Substitute both values back:1) For \( y = 13 \): \[ x^2 - 5x - 1 = 13 \] \[ x^2 - 5x - 14 = 0 \]2) For \( y = 5 \): \[ x^2 - 5x - 1 = 5 \] \[ x^2 - 5x - 6 = 0 \]
05

- Solve the Quadratic Equations

Solve the two quadratic equations:1) \( x^2 - 5x - 14 = 0 \):Using the quadratic formula, solve for \( x \): \[ x = \frac{5 \pm \sqrt{25 + 56}}{2} \] \[ x = \frac{5 \pm \sqrt{81}}{2} \] \[ x = \frac{5 \pm 9}{2} \] \[ x = 7 \text{ or } x = -2 \]2) \( x^2 - 5x - 6 = 0 \):Using the quadratic formula, solve for \( x \): \[ x = \frac{5 \pm \sqrt{25 + 24}}{2} \] \[ x = \frac{5 \pm \sqrt{49}}{2} \] \[ x = \frac{5 \pm 7}{2} \] \[ x = 6 \text{ or } x = -1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic formula
The quadratic formula is a powerful tool for solving equations of the form \( ax^2 + bx + c = 0 \). It states that the solutions for x can be found using: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a \) represents the coefficient of \( x^2 \), \( b \) represents the coefficient of \( x \), and \( c \) is the constant term. This formula helps in finding the values of \( x \) for any quadratic equation. In our exercise, we used the quadratic formula to solve for \( y \) in a simpler quadratic equation after we performed a substitution.
substitution method
The substitution method is used to simplify complex equations, making them easier to solve. In our exercise, we substituted \( y = x^2 - 5x - 1 \). This changed the original equation \( (x^2 - 5x - 1)^2 - 18(x^2 - 5x - 1) + 65 = 0 \) into a simpler quadratic equation \( y^2 - 18y + 65 = 0 \). This allowed us to use the quadratic formula to solve for \( y \). By simplifying the problem, we reduced the difficulty and could focus on solving a straightforward quadratic equation.
back-substitution
Back-substitution is the final step where we revert our substitution to find the actual values of the original variable. After determining \( y = 13 \) and \( y = 5 \), we replaced \( y \) back with \( x^2 - 5x - 1 \). This resulted in two equations: \( x^2 - 5x - 14 = 0 \) and \( x^2 - 5x - 6 = 0 \). Solving these quadratic equations using the quadratic formula or factoring, we found the final values of \( x \). This step is crucial to translate our intermediate solutions back to the context of the original problem.

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Most popular questions from this chapter

Solve. $$ \left(t+\frac{3}{2}\right)^{2}=\frac{7}{2} $$

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