Solve. Check for extraneous solutions.
1) (/3x+7) = x-1
2) (3/2x)-3=9
Note: (/3x+7) is the square root of 3x plus 7 and (3/2x)-3=9 is 3 times the square root of 2x - 3 equals 9.
Let's solve the equations step by step:
1) (√(3x + 7)) = x - 1
Step 1: Square both sides of the equation to eliminate the square root.
(√(3x + 7))^2 = (x - 1)^2
Step 2: Simplify.
3x + 7 = x^2 - 2x + 1
Step 3: Rearrange the equation in quadratic form.
x^2 - 2x + 1 - 3x - 7 = 0
x^2 - 5x - 6 = 0
Step 4: Factor the quadratic equation.
(x - 6)(x + 1) = 0
Step 5: Set each factor equal to zero and solve for x.
x - 6 = 0 or x + 1 = 0
x = 6 or x = -1
Now, let's check for extraneous solutions by substituting the found solutions back into the original equation:
For x = 6:
√(3(6) + 7) = 6 - 1
√(25) = 5
5 = 5 (True)
For x = -1:
√(3(-1) + 7) = -1 - 1
√(4) = -2
2 = -2 (False)
Therefore, the only solution that satisfies the original equation is x = 6. x = -1 is an extraneous solution.
2) (3√(2x - 3)) - 3 = 9
Step 1: Add 3 to both sides of the equation.
(3√(2x - 3)) = 12
Step 2: Divide both sides of the equation by 3 to isolate the square root.
√(2x - 3) = 4
Step 3: Square both sides of the equation to eliminate the square root.
(√(2x - 3))^2 = 4^2
Step 4: Simplify.
2x - 3 = 16
Step 5: Add 3 to both sides of the equation.
2x = 19
Step 6: Divide both sides of the equation by 2 to solve for x.
x = 19/2
Now, let's check for extraneous solutions by substituting the found solution back into the original equation:
For x = 19/2:
(3√(2(19/2) - 3)) - 3 = 9
(3√(19 - 3)) - 3 = 9
(3√16) - 3 = 9
(3(4)) - 3 = 9
12 - 3 = 9
9 = 9 (True)
The solution x = 19/2 satisfies the original equation.
To solve each equation and check for extraneous solutions, we will follow these steps:
Step 1: Isolate the radical term on one side of the equation.
Step 2: Square both sides of the equation.
Step 3: Solve the resulting equation.
Step 4: Check for extraneous solutions by substituting the obtained solutions back into the original equation.
Let's start with equation 1:
1) (√(3x+7)) = x - 1
Step 1: Move x to the left side, and isolate the radical term:
√(3x + 7) = x - 1
Step 2: Square both sides:
(√(3x + 7))^2 = (x - 1)^2
(3x + 7) = (x - 1)^2
Step 3: Expand the right side and simplify:
3x + 7 = x^2 - 2x + 1
Step 4: Set the equation to zero:
0 = x^2 - 5x - 6
Step 5: Factor the quadratic equation:
0 = (x - 6)(x + 1)
Step 6: Set each factor equal to zero and solve:
x - 6 = 0 or x + 1 = 0
x = 6 or x = -1
To check for extraneous solutions, substitute each solution back into the original equation:
For x = 6:
√(3(6) + 7) = 6 - 1
√(18 + 7) = 5
√25 = 5
5 = 5 (True)
For x = -1:
√(3(-1) + 7) = -1 - 1
√(3 - 7) = -2
√-4 = -2
No real solution exists for x = -1.
Therefore, the only valid solution to equation 1 is x = 6. No extraneous solution exists.
Now, let's move on to equation 2:
2) 3√(2x - 3) = 9
Step 1: Isolate the radical term:
√(2x - 3) = 9/3
√(2x - 3) = 3
Step 2: Square both sides:
(√(2x - 3))^2 = 3^2
2x - 3 = 9
Step 3: Solve for x:
2x = 9 + 3
2x = 12
x = 6
To check for extraneous solutions, substitute the solution back into the original equation:
√(2(6) - 3) = 3
√(12 - 3) = 3
√9 = 3
3 = 3 (True)
Therefore, the valid solution to equation 2 is x = 6. No extraneous solution exists.
Final solutions:
1) x = 6
2) x = 6