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Chapter 3: Problem 59

Solve. $$\sqrt{x+7}=x+1$$

Short Answer

Expert verified
The solution is \ x = 2 \

Step by step solution

01

- Isolate the Radical Expression

The radical expression \(\root{x+7}\) is already isolated on one side of the equation: \[ \root{x+7} = x+1 \].
02

- Square Both Sides

Square both sides of the equation to eliminate the radical: \[ (\root{x+7})^2 = (x+1)^2 \] This simplifies to: \[ x+7 = (x+1)(x+1) \] Expand the right side: \[ x+7 = x^2 + 2x + 1 \]}, {
03

- Rearrange into a Standard Quadratic Equation

Move all terms to one side of the equation to set the quadratic expression to zero: \[ x+7 - (x^2 + 2x + 1) = 0 \] Combine like terms: \[ -x^2 - x + 6 = 0 \] Multiply the entire equation by -1 to simplify: \[ x^2 + x - 6 = 0 \]
04

- Factor the Quadratic Equation

Factor the quadratic equation \ x^2 + x - 6 \ : \[ (x + 3)(x - 2) = 0 \]
05

- Find the Solutions

Set each factor equal to zero and solve for x: \[ x + 3 = 0 \] \[ x = -3 \] \[ x - 2 = 0 \] \[ x = 2 \]
06

- Verify the Solutions

Substitute both solutions back into the original equation \ \root{x+7} = x + 1 \ to verify: \[ x = -3 \] \[ \root{-3 + 7} = -3 + 1 \] \[ \root{4} = -2 \] which is false. Therefore, \ x = -3 \ is not a solution. Verifying \ x = 2 \ : \[ \root{2 + 7} = 2 + 1 \] \[ \root{9} = 3 \] which is true. Therefore, \ x = 2 \ is a solution.

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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 type of polynomial equation that follow the standard form: \[ ax^2 + bx + c = 0 \]Here,
  • a, b,
  • and c are constants,
  • and x represents the variable.
The 'a' term must be non-zero, otherwise, the equation becomes linear. A quadratic equation typically has two solutions, found by methods like factoring, completing the square, or using the quadratic formula.

In the given exercise, after squaring both sides, we obtain a quadratic equation from \[ x + 7 = x^2 + 2x + 1.\]Rearranging the terms, we get \[ -x^2 - x + 6 = 0. \] Multiplying the entire equation by -1 simplifies it to the standard quadratic form \[ x^2 + x - 6 = 0. \]
Factoring
Factoring is a method used to solve quadratic equations by expressing the quadratic as a product of its binomial factors. Consider \[ x^2 + x - 6 = 0.\]We look for two numbers that multiply to -6 (the constant term) and add up to 1 (the coefficient of x).

These numbers are +3 and -2 because
  • 3 * -2 = -6,
  • 3 + (-2) = 1.
Thus, we can factor the quadratic equation as\[ (x + 3)(x - 2) = 0.\]
Setting each factor equal to zero gives the solutions:\[ x + 3 = 0 \rightarrow x = -3, \]\[ x - 2 = 0 \rightarrow x = 2. \]
Verification of Solutions
Verifying solutions ensures the values found actually satisfy the original equation. Even if algebraic manipulation says a value is a solution, it must still work in the context of the original problem.

Here, we substitute the solutions back into the original radical equation \[ \root{x+7} = x + 1. \]
  • For x = -3:

    • \[ \root{-3 + 7} = -3 + 1 \ \rightarrow \root{4} eq -2 \]
    which is false. Therefore, x = -3 is not a solution.
  • For x = 2:

    • \[ \root{2 + 7} = 2 + 1 \ \rightarrow \root{9} = 3 \ \rightarrow \text{which is true}. \]
Thus, x = 2 is a verified solution.

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Most popular questions from this chapter

Solve and write interval notation for the solution set. Then graph the solution set. $$|3 x+5|<0$$

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Solve. $$t^{1 / 5}=2$$

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