I would appreciate your help with this.
Thank you
Solve each equation and check for extraneous solutions.
[SQRT(x - 3)] - 7=0
2[SQRT(w + 4)]=5
[SQRT(x)]=-2
SQRT(x - 3) - 7 = 0
Change it to
SQRT(x - 3) = 7
and then square both sides, to get
x - 3 = 49
x = 52
x must be >3 and sqrt (x-3) must be positive (since is equals 7), so there are no other solutions.
Square both sides of the second equation and proceed similarly
What would i do with the 2 ?
Square it, or first divide both sides of the equation by 2 and then square 5/2 on the right side.
To solve each equation and check for extraneous solutions, follow these steps:
1. [SQRT(x - 3)] - 7 = 0
Add 7 to both sides:
[SQRT(x - 3)] = 7
Square both sides:
(x - 3) = 49
Add 3 to both sides:
x = 52
However, we need to check if x = 52 is an extraneous solution because we squared both sides of the equation. Plugging the value of x = 52 back into the original equation, [SQRT(52 - 3)] - 7 = 0:
[SQRT(49)] - 7 = 0
7 - 7 = 0
0 = 0
Since the left side of the equation is equal to the right side, x = 52 is not an extraneous solution.
2. 2[SQRT(w + 4)] = 5
Divide both sides by 2:
[SQRT(w + 4)] = 2.5
Square both sides:
w + 4 = 6.25
Subtract 4 from both sides:
w = 2.25
Check for extraneous solution by plugging w = 2.25 back into the original equation: 2[SQRT(2.25 + 4)] = 5:
2[SQRT(6.25)] = 5
2(2.5) = 5
5 = 5
Since the left side is equal to the right side, w = 2.25 is not an extraneous solution.
3. [SQRT(x)] = -2
Since the square root of a number cannot be negative, there are no solutions for this equation.