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Find the extraneous solution for the radical equations given below.
Problem 1 :
√(x + 4) = x - 2 Solution
Problem 2 :
√(10x + 9) = x + 3 Solution
Problem 3 :
√(2x + 5) = √(x + 7) Solution
Problem 4 :
4) √(x + 6) – 2 = √(x – 2) Solution
Problem 5 :
In a hurricane, the mean sustained wind velocity v (in meters per second) can be modeled by v( p) = 6.3 √(1013 − p), where p is the air pressure (in millibars) at the center of the hurricane. Estimate the air pressure at the center of the hurricane when the mean sustained wind velocity is 54.5 meters per second.
Problem 6 :
Biologists have discovered that the shoulder height h (in centimeters) of a male Asian elephant can be modeled by h = 62.5 3√t + 75.8, where t is the age (in years) of the elephant. Determine the age of an elephant with a shoulder height of 250 centimeters
1) 5 is extraneous solution.
2) there is no extraneous solution.
3) there is no extraneous solution.
4) there is no extraneous solution.
5) The air pressure at the center of the hurricane is about 938 millibars.
6) Approximately 22 years.
Solve the following radical equations with rational exponents.
Problem 1 :
2x(2/3) = 8
Problem 2 :
4x(3/2) = 32
Problem 3 :
x1/4 + 3 = 0
Problem 4 :
2x3/4 - 14 = 40
Problem 5 :
(x + 6)1/2 = x
Problem 6 :
(5 - x)1/2 – 2x = 0
Problem 7 :
2(x + 11)1/2 = x + 3
Problem 8 :
(5x2 – 4)1/4 = x
Problem 9 :
x1/5 = 2
Problem 10 :
2√(3x – 1) + 3 = 11
Problem 11 :
4x2 = 64
Problem 12 :
2(x – 2)4 – 3 = 159
Problem 13 :
√(2x + 4) = √(x + 2)
Problem 14 :
∛x – 6 = -2
Problem 15 :
3 = (2x + 7)1/4
Problem 16 :
Let f(x) = 6x and g(x) = x3/4. Find ( f/g ) (x), then evaluate the quotient when x = 16.
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1) x = -8 and x = 8 2) x = 4 3) x = 81 4) x = 81 5) x = 3 and x = -2 6) x = 1 and x = -5/4 7) x = 5 and x = -7 8) x = ±1 and x = ±2 |
9) x = 32 10) x = 17//3 11) x = ±4 12) x = 5 and x = -1 13) x = -2 14) x = 64 15) x = 37 16) 12 |
Solve the following radical equations :
Problem 1 :
√(x + 1) = 4
Problem 2 :
√(2t + 15) = t
Problem 3 :
√(p - 4) = -5
Problem 4 :
√(2 – x) = x - 2
Problem 5 :
√s2 – 1 = √5s – 5
Problem 6 :
∛x = 4
Problem 7 :
∜(2p + 2) = 3
Problem 8 :
∛x + 1 = ∛x2 - 5
Problem 9 :
√(2 - √x) = √x
Problem 10 :
√(3a + 16) - 3 = a - 1
In the equation above, a ≥ 0, which of the following is a possible value of a ?
a) 3 b) 2 c) 1 d) 0
Problem 11 :
8 + [√(2x + 29)/3] = 9
For what value of x is this equation true ?
a) -10 b) -12 c) 19 d) No solution
Problem 12 :
3x = x + 14
√(3z2 - 11) + 2x = 22
If z > 0, what is the value of z ?
1) x = 15
2) t = 5 and t = -3
3) p = 29
4) x = 1 and x = 2
5) s = 1 and s = 4
6) x = 64
7) p = 39.5
8) x = -2 and x = 3
9) x = 1 and x = 4
10) a = -4 and a = 3
11) x = -10
12) z = -5 and 5
Solve the following radical equations.
Problem 1 :
3√x + 3 = 15
Problem 2 :
4√x - 1 = 3
Problem 3 :
√(x + 3) = 5
Problem 4 :
√(3x + 4) = 4
Problem 5 :
√(2x + 3) - 7 = 0
Problem 6 :
√(6 - 3x) - 2 = 0
Problem 7 :
x1/4 - 1 = 0
Problem 8 :
(x - 2)1/3 = -5
Problem 9 :
x1/3 - 2 = 0
Problem 10 :
√3x = 6
Problem 11 :
(2x + 7)1/2 - x = 2
Problem 12 :
√4x - 8 = 0
Problem 13 :
In an amusement park ride, a rider suspended by cables swings back and forth from a tower. The maximum speed v (in meters per second) of the rider can be approximated by v = √2gh, where h is the height (in meters) at the top of each swing and g is the acceleration due to gravity (g ≈ 9.8 m/sec2). Determine the height at the top of the swing of a rider whose maximum speed is 15 meters per second.
Problem 14 :
(5x2 − 4)1/4 = x
Problem 15 :
2(x + 11)1/2 = x + 3
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1) x = 16 2) x = 1 3) x = 22 4) x = 4 5) x = 23 6) x = 2/3 7) x = 1 8) x = -123 |
9) x = 8 10) x = 12 11) x = 1, -3 12) x = 16 13) h = 11.48 meters 14) x = -1 and 1, x = -2 and 2 15) x = -7 and x = 5 |
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May 21, 24 08:51 PM
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