PROPERTIES OF SQUARE ROOTS
1. If two or more numbers multiplied with square roots, you can take the square root once and multiply the numbers inside the square root.
√m x √n = √(m x n)
2. If two numbers are in division with square roots, you can take the square root once and divide the numbers inside the square root.
√m/√n = √(m/n)
3. One number can be taken out of the square root for every two same numbers multiplied inside the square root.
√(m x m) = m
√4 = √(2 x 2) = 2
4. Square root of a number can be written as exponent 1/2.
√m = m1/2
5. Addition and subtraction of two or more numbers with square roots can be performed with like radicands only. (Radicand is the number inside the square root)
For example, 9√2 and 4√2 can be added or subtracted. Because the numbers inside the square roots are same.
9√2 + 4√2 = 13√2
9√2 - 4√2 = 5√2
6. If a square root is moved from side of the equation to the other side, it will become square.
√x = a
x = a2
7. If a square is moved from side of the equation to the other side, it will become square root.
y2 = b
y = √b
8. If the digit in one's place of a number is 2, 3, 7 or 8, then the number can not be a perfect square. So the square root of such numbers will be irrational.
For example, √23 = 4.795831.........
9. If a number ends with odd number of zeros, then, the square root of the number will be irrational.
For example, √3000 = 54.772255.......
10. The square of a perfect square is always a rational number.
√4 = √(2 x 2) = 2
√49 = (7 x 7) = 7
11. The square root of an even perfect square number is always even and the square root of an odd perfect square number is always is odd.
For example,
√144 = 144
√225 = 15
12. Square root of a negative number is considered to be an imaginary value.
For example, √(-9), √(-12).
Solved Problems
Problem 1 :
Simplify :
√6 ⋅ √3
Solution :
= √6 ⋅ √3
= √(6 ⋅ 3)
= √(2 ⋅ 3 ⋅ 3)
= 3√2
Problem 2 :
Simplify :
√35 ÷ √7
Solution :
= √35 ÷ √7
= √(35/7)
= √5
Problem 3 :
Simplify :
3√425 + 4√68
Solution :
Decompose 425 and 68 into prime factors using synthetic division.
|
√425 = √(5 ⋅ 5 ⋅ 17) √425 = 5√17 |
√68 = √(2 ⋅ 2 ⋅ 17) √68 = 2√17 |
3√425 + 4√68 :
= 3(5√17) + 4(2√17)
= 15√17 + 8√17
= 23√17
Problem 4 :
Simplify :
√243 - 5√12 + √27
Solution :
Decompose 243, 12 and 27 into prime factors using synthetic division.
√243 = √(3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3) = 9√3
√12 = √(2 ⋅ 2 ⋅ 3) = 2√3
√27 = √(3 ⋅ 3 ⋅ 3) = 3√3
√243 - 5√12 + √27 :
= 9√3 - 5(2√3) + 3√3
= 9√3 - 10√3 + 3√3
= 2√3
Problem 5 :
If √a = 1/5, then find the value of a.
Solution :
√a = 1/5
a = (1/5)2
a = 12/52
a = 1/25
Problem 6 :
If (√4)7 ⋅ (√2)-4 = 2k, then solve for k.
Solution :
(√4)7 ⋅ (√2)-4 = 2k
27 ⋅ (21/2)-4 = 2k
27 ⋅ 2-2 = 2k
27 - 2 = 2k
25 = 2k
k = 5
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