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Chapter 3: Problem 11

Find all real solutions to each equation. Check your answers. $$\sqrt{x+1}=x-5$$

Short Answer

Expert verified
The only real solution is x = 8.

Step by step solution

01

- Isolate the square root

Given the equation \(\root{x+1}=x-5\), notice that the square root term is already isolated.
02

- Square both sides

Square both sides of the equation to remove the square root. \[ (\root{x+1})^2 = (x-5)^2 \] This simplifies to: \[ x + 1 = (x-5)^2 \]
03

- Expand and simplify

Expand the right side of the equation. \[ x + 1 = x^2 - 10x + 25 \] Next, move all terms to one side to set the equation to zero. \[ 0 = x^2 - 11x + 24 \]
04

- Solve the quadratic equation

The quadratic equation \(0 = x^2 - 11x + 24\) can be solved by factoring. \[ x^2 - 11x + 24 = (x-3)(x-8) \] Set each factor to zero and solve for x. \[ x-3=0 \] \[ x = 3 \] \[ x-8=0 \] \[ x = 8 \]
05

- Check the solutions

Substitute each solution back into the original equation to verify.For \( x = 3 \): \[ \root{3+1} = 3-5 \] \[ 2 eq -2 \] (False, so x = 3 is not a solution)For \( x = 8 \): \[ \root{8+1} = 8-5 \] \[ 3 = 3 \] (True, so x = 8 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
A quadratic equation is a second-degree polynomial equation that has the form \[ ax^2 + bx + c = 0 \].
In this exercise, we arrive at the quadratic equation \[ x^2 - 11x + 24 = 0 \] after squaring both sides of the original equation and simplifying.
Quadratic equations can have two solutions, one solution, or no real solutions depending on the value of the discriminant \( D = b^2 - 4ac \). If \( D > 0 \), there are two real solutions. If \( D = 0 \), there is one real solution. If \( D < 0 \), there are no real solutions.
In our case, the quadratic equation \( x^2 - 11x + 24 = 0 \) was solved by factoring.
Factoring
Factoring is the process of breaking down a polynomial into simpler polynomials whose product is the original polynomial.
To factor the quadratic equation \( x^2 - 11x + 24 \), we look for two numbers that multiply to give \( 24 \) (the constant term) and add to give \( -11 \) (the linear coefficient).
These numbers are \( -3 \) and \( -8 \). Thus, we write:
\( x^2 - 11x + 24 = (x-3)(x-8) \).
After factoring, we set each factor equal to zero:
\( x-3 = 0 \), which gives \( x = 3 \), and \( x-8 = 0 \), which gives \( x = 8 \).
These are the potential solutions to the quadratic equation.
Checking Solutions
After obtaining the potential solutions \( x = 3 \) and \( x = 8 \), it's crucial to check whether these solutions satisfy the original equation \( \sqrt{x+1} = x-5 \).
Substituting \( x = 3 \) into the original equation gives:
\( \sqrt{3 + 1} = 3 - 5 \)
\( \sqrt{4} = -2 \)
\( 2 eq -2 \), so \( x = 3 \) is not a valid solution.
Substituting \( x = 8 \) into the original equation gives:
\( \sqrt{8 + 1} = 8 - 5 \)
\( \sqrt{9} = 3 \)
\( 3 = 3 \), so \( x = 8 \) is a valid solution.
Always remember to check your solutions in the original equation, especially when dealing with radical equations, as extraneous solutions can arise during the process.

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Most popular questions from this chapter

Use division to write each rational expression in the form quotient \(+\) remainder/divisor. Use synthetic division when possible. $$\frac{x-1}{x+3}$$

Use division to write each rational expression in the form quotient \(+\) remainder/divisor. Use synthetic division when possible. $$\frac{4 t-5}{2 t+1}$$

For the function \(y=-4 x^{2}+8 x+9\) find the domain, range, and the maximum value for \(y\)

Given an arbitrary rational function \(R(x),\) explain how you can find the approximate value of \(R(x)\) when the absolute value of \(x\) is very large.

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