Questions: Solve. Check for extraneous solutions. sqrt(8x+33)=x+6

Transcript text: Solve. Check for extraneous solutions. \[ \sqrt{8 x+33}=x+6 \]

Solution

Solution Steps

Step 1: Set Up the Equation

We start with the equation \[ \sqrt{8x + 33} = x + 6. \]

Step 2: Square Both Sides

To eliminate the square root, we square both sides: \[ 8x + 33 = (x + 6)^2. \]

Step 3: Expand and Rearrange

Expanding the right side gives us: \[ 8x + 33 = x^2 + 12x + 36. \] Rearranging this equation leads to: \[ 0 = x^2 + 12x + 36 - 8x - 33, \] which simplifies to: \[ 0 = x^2 + 4x + 3. \]

Step 4: Factor the Quadratic

We can factor the quadratic equation: \[ 0 = (x + 3)(x + 1). \]

Step 5: Solve for \( x \)

Setting each factor to zero gives us the solutions: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3, \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1. \]

Step 6: Check for Extraneous Solutions

We substitute \( x = -3 \) and \( x = -1 \) back into the original equation to check for extraneous solutions.

For \( x = -3 \): \[ \sqrt{8(-3) + 33} = \sqrt{-24 + 33} = \sqrt{9} = 3, \] and \[ -3 + 6 = 3. \] Both sides are equal, so \( x = -3 \) is valid.

For \( x = -1 \): \[ \sqrt{8(-1) + 33} = \sqrt{-8 + 33} = \sqrt{25} = 5, \] and \[ -1 + 6 = 5. \] Both sides are equal, so \( x = -1 \) is also valid.

Step 7: Conclusion

The solutions to the equation \(\sqrt{8x + 33} = x + 6\) are \( x = -3 \) and \( x = -1\), and both solutions are valid with no extraneous solutions.