√x+1 -√9-x= √2x-12
Let's analyze and solve the equation step by step:
Given: \[ \sqrt{x + 1} - \sqrt{9 - x} = \sqrt{2x - 12} \]
Step 1: Determine the domain
- For \(\sqrt{x+1}\), we need \(x + 1 \geq 0\) \(\Rightarrow x \geq -1\).
- For \(\sqrt{9 - x}\), we need \(9 - x \geq 0\) \(\Rightarrow x \leq 9\).
- For \(\sqrt{2x - 12}\), we need \(2x - 12 \geq 0\) \(\Rightarrow x \geq 6\).
Combining these: \[ x \geq 6 \quad \text{and} \quad x \leq 9 \] So, the domain is \([6, 9]\).
Step 2: Isolate and manipulate the equation
Start with: \[ \sqrt{x + 1} - \sqrt{9 - x} = \sqrt{2x - 12} \]
Let's move all square roots to one side: \[ \sqrt{x + 1} = \sqrt{9 - x} + \sqrt{2x - 12} \]
Step 3: Square both sides to eliminate the square root
\[ (\sqrt{x + 1})^2 = (\sqrt{9 - x} + \sqrt{2x - 12})^2 \]
Calculate: \[ x + 1 = (9 - x) + (2x - 12) + 2 \sqrt{(9 - x)(2x - 12)} \] Simplify: \[ x + 1 = 9 - x + 2x - 12 + 2 \sqrt{(9 - x)(2x - 12)} \] \[ x + 1 = (9 - 12) + (-x + 2x) + 2 \sqrt{(9 - x)(2x - 12)} \] \[ x + 1 = -3 + x + 2 \sqrt{(9 - x)(2x - 12)} \]
Subtract \(x\) from both sides: \[ 1 = -3 + 2 \sqrt{(9 - x)(2x - 12)} \]
Add 3 to both sides: \[ 4 = 2 \sqrt{(9 - x)(2x - 12)} \]
Divide both sides by 2: \[ 2 = \sqrt{(9 - x)(2x - 12)} \]
Now, square both sides again: \[ (2)^2 = (9 - x)(2x - 12) \]
Calculate: \[ 4 = (9 - x)(2x - 12) \]
Step 4: Expand and form a quadratic
Expand the right side: \[ 4 = (9)(2x - 12) - x(2x - 12) \] \[ 4 = 18x - 108 - 2x^2 + 12x \]
Simplify: \[ 4 = (18x + 12x) - 108 - 2x^2 \] \[ 4 = 30x - 108 - 2x^2 \]
Bring all to one side: \[ 2x^2 - 30x + 112 = 0 \]
Divide through by 2: \[ x^2 - 15x + 56 = 0 \]
Step 5: Solve the quadratic
Factors of 56 that sum to 15: \[ x^2 - 15x + 56 = 0 \]
Factor: \[ (x - 7)(x - 8) = 0 \]
Solutions: \[ x = 7 \quad \text{or} \quad x = 8 \]
Recall the domain: \([6, 9]\). Both are within the domain.
Step 6: Verify solutions in the original equation
Check \(x=7\):
Left side: \[ \sqrt{7 + 1} - \sqrt{9 - 7} = \sqrt{8} - \sqrt{2} \approx 2.828 - 1.414 = 1.414 \]
Right side: \[ \sqrt{2 \times 7 - 12} = \sqrt{14 - 12} = \sqrt{2} \approx 1.414 \]
Equal, so \(x=7\) is a solution.
Check \(x=8\):
Left side: \[ \sqrt{8 + 1} - \sqrt{9 - 8} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2 \]
Right side: \[ \sqrt{2 \times 8 - 12} = \sqrt{16 - 12} = \sqrt{4} = 2 \]
Equal, so \(x=8\) is a solution.
Final answer:
\[ \boxed{ \text{Solutions are } x=7 \quad \text{and} \quad x=8 } \]
both within the valid domain and satisfying the original equation.
