Problem 40 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\sqrt{6 x+7}-9=x-7$$

Step-by-Step Solution

Answer

x = 3, -1

Step 1: Isolate the Square Root

To start, isolate the square root on one side of the equation. Add 9 to both sides of the equation to get: \[\sqrt{6x + 7} = x + 2\]

Step 2: Square Both Sides

Next, eliminate the square root by squaring both sides of the equation: \[(\sqrt{6x + 7})^2 = (x + 2)^2\] \[6x + 7 = x^2 + 4x + 4\]

Step 3: Rearrange the Equation

Move all terms to one side to set the equation to 0: \[0 = x^2 - 2x - 3\] This is a standard quadratic equation.

Step 4: Factor the Quadratic Equation

Factor the quadratic equation: \[(x - 3)(x + 1) = 0\]

Step 5: Solve for x

Set each factor equal to zero and solve for x: \[ x - 3 = 0 \rightarrow x = 3\] \[ x + 1 = 0 \rightarrow x = -1\]

Step 6: Check for Extraneous Solutions

Verify both solutions in the original equation. \[\sqrt{6(3) + 7} - 9 = 3 - 7\] \[\sqrt{18 + 7} - 9 = -4\] \[\sqrt{25} - 9 = -4\] \[5 - 9 = -4\] (True) \[\sqrt{6(-1) + 7} - 9 = -1 - 7\] \[\sqrt{-6 + 7} - 9 = -8\] \[\sqrt{1} - 9 = -8\] \[1 - 9 = -8\] (True) Both solutions are valid.

Key Concepts

Isolating Square Roots

Isolating the square root is a crucial first step in solving equations that involve square roots. The goal here is to get the square root term by itself on one side of the equation. So, if you start with an equation like \( \sqrt{6x + 7} - 9 = x - 7 \), you need to manipulate it to isolate the square root. You can do this by performing operations on both sides to balance the equation.

For instance, adding 9 to both sides helps us to isolate the square root: \[ \sqrt{6x + 7} = x + 2 \]. This isolation makes it easier to deal with the square root in the next steps. Isolating the square root simplifies the problem and allows us to proceed to squaring both sides, which eliminates the square root.

Factoring Quadratics

Once you have isolated the square root and squared both sides of the equation, you end up with a quadratic equation. For example, after squaring, the equation \[ \sqrt{6x + 7} = x + 2 \] becomes \[ 6x + 7 = x^2 + 4x + 4 \].

The next step is to rearrange this equation to set it equal to zero: \[ x^2 - 2x - 3 = 0 \]. This is now in the standard form for a quadratic equation. To solve it, you can use factoring.

Factoring involves expressing the quadratic equation as a product of its factors. For the equation \[ 0 = x^2 - 2x - 3 \], we can factor it into: \[ (x - 3)(x + 1) = 0 \]. Each factor set to zero gives potential solutions for \text{x}\. So, \[ x - 3 = 0 \rightarrow x = 3 \] and \[ x + 1 = 0 \rightarrow x = -1 \].

Checking Extraneous Solutions

After solving the quadratic equation through factoring, you obtain potential solutions. It is essential to check these solutions to ensure they work in the original equation and are not extraneous. Extraneous solutions are results that are mathematically correct for the transformed equation but do not satisfy the original equation.

For instance, with potential solutions \( x = 3 \) and \( x = -1 \), plug them back into the original equation \( \sqrt{6x + 7} - 9 = x - 7 \).

For \( x = 3 \):
\[ \sqrt{6(3) + 7} - 9 = 3 - 7 \]
\[ \sqrt{18 + 7} - 9 = -4 \]
\[ \sqrt{25} - 9 = -4 \]
\[ 5 - 9 = -4 \] (True)

For \( x = -1 \):
\[ \sqrt{6(-1) + 7} - 9 = -1 - 7 \]
\[ \sqrt{-6 + 7} - 9 = -8 \]
\[ \sqrt{1} - 9 = -8 \]
\[ 1 - 9 = -8 \] (True)

Both solutions, \( x = 3 \) and \( x = -1 \), are valid as they satisfy the original equation. This step is vital in verifying your solutions are correct and applicable.