Problem 1 :
Solve for x :
x2 = 121
Problem 2 :
Solve for x :
x2 = 16/169
Problem 3 :
Solve for x :
x3 = 729
Problem 4 :
Solve for x :
x3 = 8/125
Problem 5 :
Find the square root of 0.16.
Problem 6 :
Find the cube root of 0.008.
Problem 7 :
Find the cube root of -125.
Problem 8 :
Can you solve the equation x2 = 27 and get a rational umber as solution ? Explain.
Problem 1 :
Solve for x :
x2 = 121
Solution :
x2 = 121
Solve for x by taking square root on both sides.
x = ± √121
Apply the definition of square root.
Think: What numbers squared equal 121 ?
x = ± 11
So, the solutions are 11 and −11.
Problem 2 :
Solve for x :
x2 = 16/169
Solution :
x2 = 16/169
Solve for x by taking square root on both sides.
x = ± √(16/169)
Apply the definition of square root.
Think: What numbers squared equal 16/169 ?
x = ± 4/13
So, the solutions are 4/13 and −4/13.
Problem 3 :
Solve for x :
x3 = 729
Solution :
x3 = 729
Solve for x by taking cube root on both sides.
x = 3√729
Apply the definition of cube root.
Think: What number cubed equals 729 ?
x = 9
So, the solution is 9.
Problem 4 :
Solve for x :
x3 = 8/125
Solution :
x3 = 8/125
Solve for x by taking cube root on both sides.
x = 3√(8/125)
Apply the definition of cube root.
Think: What number cubed equals 8/125 ?
x = 2/5
So, the solution is 2/5.
Problem 5 :
Find the square root of 0.16.
Solution :
√0.16 = √(16/100)
=√16/√100
= 4/10
= 0.4
Problem 6 :
Find the cube root of 0.008.
Solution :
3√0.008 = 3√(8/1000)
=3√8/3√1000
= 2/10
= 0.2
Solve for x by taking cube root on both sides.
x = 3√(8/125)
Apply the definition of cube root.
Think: What number cubed equals 8/125 ?
x = 2/5
So, the solution is 2/5.
Problem 7 :
Find the cube root of -125.
Solution :
3√-125 = 3√(-5 x -5 x -5)
= -5
Problem 8 :
Can you solve the equation x2 = 27 and get a rational umber as solution ? Explain.
Solution :
No, we can not get a rational number as a solution.
In the given equation we have "square" for "x".
So, we will be able to get a rational number as solution, only if we have a perfect square on the right side of the equation.
Since we have 27 (not a perfect square) on the right side of the equation, we can not get a rational number as a solution.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Apr 23, 24 09:10 PM
Apr 23, 24 12:32 PM
Apr 23, 24 12:07 PM