i need help solving this one (x + 4)^2/3= 9 a) 13 b) 23;-31 c) 5 d) 31;-31
also sove check for extraneous solutions
(6x-5)^1/2 = (7+ 4x)^1/2
a) 1/6 b) -9 c) 6 or d) 1/5
(x + 4)^2/3= 9
[(x + 4)^2/3]^(3/2)= 9^(3/2)
x+4 = 27
x = 23
do the second one by squaring both sides
ok so for this one i need help solving this one (x + 4)^2/3= 9 a) 13 b) 23;-31 c) 5 d) 31;-31
it would be c 23 or -31 right?
for for the 2nd one i got -9?
but i don't think i did it right
To solve the equation (x + 4)^(2/3) = 9, we'll follow these steps:
Step 1: Isolate the exponent
Raise both sides of the equation to the reciprocal of the exponent, which is (3/2). This will eliminate the exponent on the left side, giving us:
[(x + 4)^(2/3)]^(3/2) = 9^(3/2)
Simplifying:
[(x + 4)^(2/3 * 3/2)] = 9^(3/2)
(x + 4)^1 = 9^(3/2)
Step 2: Simplify
Since raising to the power of 1 does not change the base, we can simplify the left side to just (x + 4):
x + 4 = 9^(3/2)
Step 3: Find the square root
Take the square root of both sides of the equation to solve for x:
√(x + 4) = √(9^(3/2))
Simplifying using the rules of exponents:
√(x + 4) = (9^(1/2))^(3/2)
√(x + 4) = 3^3
√(x + 4) = 27
Step 4: Solve for x
Square both sides of the equation to eliminate the square root:
(x + 4) = 27^2
x + 4 = 729
x = 729 - 4
x = 725
So, the solution to the equation (x + 4)^(2/3) = 9 is x = 725.
Now let's move on to the second equation:
To solve the equation (6x - 5)^(1/2) = (7 + 4x)^(1/2), we'll follow these steps:
Step 1: Square both sides of the equation
Raise both sides of the equation to the power of 2 to eliminate the square root:
[(6x - 5)^(1/2)]^2 = [(7 + 4x)^(1/2)]^2
Simplifying:
6x - 5 = 7 + 4x
Step 2: Solve for x
Subtract 4x from both sides and add 5 to both sides:
6x - 4x = 7 + 5
2x = 12
x = 12/2
x = 6
So, the solution to the equation (6x - 5)^(1/2) = (7 + 4x)^(1/2) is x = 6.
Now, to check for extraneous solutions, substitute the found value of x back into each equation and see if it satisfies the original equations.
For the first equation, substitute x = 725:
[(725 + 4)^(2/3)] = 9
[729^(2/3)] = 9
(9^2) = 9
81 = 9
Since the equation is false, x = 725 is an extraneous solution.
For the second equation, substitute x = 6:
[(6*6 - 5)^(1/2)] = (7 + 4*6)^(1/2)
[(36 - 5)^(1/2)] = (7 + 24)^(1/2)
[31^(1/2)] = (31)^(1/2)
√31 = √31
The equation is true, so x = 6 is a valid solution.