find the real solutions.
3+ the square root of (2x+2)=x
3+√(2x+2) = x
√(2x+2) = x-3
2x+2 = (x-3)^2
2x+2 = x^2-6x+9
x^2-8x+7 = 0
(x-1)(x-7) = 0
x = 1 or 7
check for extraneous roots:
3+√(2+2) = 3+2 = 5 nope
3+√16 = 3+4 = 7 ok
so, x=7 is the only solution
x=1 is the solution using the -√
3-√4 = 1
To find the real solutions to the equation 3 + √(2x+2) = x, we need to isolate the square root term and then solve for x. Here's how:
1. Start by subtracting 3 from both sides of the equation:
3 + √(2x+2) - 3 = x - 3
√(2x+2) = x - 3
2. Square both sides of the equation to eliminate the square root:
(√(2x+2))^2 = (x - 3)^2
2x + 2 = x^2 - 6x + 9
3. Move all terms to one side of the equation to form a quadratic equation:
x^2 - 6x + 9 - 2x - 2 = 0
x^2 - 8x + 7 = 0
4. Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use factoring:
(x - 7)(x - 1) = 0
5. Set each factor equal to zero and solve for x:
x - 7 = 0 or x - 1 = 0
x = 7 or x = 1
Hence, the real solutions to the equation 3 + √(2x+2) = x are x = 7 and x = 1.