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Secondary 1 | Maths
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pls help for both q
Date Posted: 3 years ago
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12. The answer is no.
As an integer, n can be negative, positive or 0.
But notice that √n² is a positive square root. So the result is a positive value.
It will work when n is a positive integer or 0.
Eg. When n = 5 , n² = 5² = 5 × 5 = 25
√n² = √25 = 5 = n
When n = 0, n² = 0² = 0 × 0 = 0
√n² = √0 = 0 = n
But if n is negative, this won't be true.
Eg. When n = -4, n² = (-4)² = (-4) × (-4) = 4 × 4 = 16
But √n² = √16 = 4 ≠ n
So the positive square root cannot be equal to a negative value of n.
Question 13 in comments.
As an integer, n can be negative, positive or 0.
But notice that √n² is a positive square root. So the result is a positive value.
It will work when n is a positive integer or 0.
Eg. When n = 5 , n² = 5² = 5 × 5 = 25
√n² = √25 = 5 = n
When n = 0, n² = 0² = 0 × 0 = 0
√n² = √0 = 0 = n
But if n is negative, this won't be true.
Eg. When n = -4, n² = (-4)² = (-4) × (-4) = 4 × 4 = 16
But √n² = √16 = 4 ≠ n
So the positive square root cannot be equal to a negative value of n.
Question 13 in comments.
13.
① Some deductions first :
n and 198/n² are both integers.
This means that n² is also an integer since the product of two integers is always an integer.
n² is positive (n² > 0) regardless of whether n is negative or positive. Because , the square of any non-zero real value is always positive.
eg. 5² = 5 × 5 = 25
eg. (-10)² = (-10) × (-10) = 10 × 10 = 100
(Note that n cannot be 0 in this question. as that would mean n² = 0² = 0 .
Then this in turn implies that 198/n² = 198/0, which is undefined since division by 0 is undefined. This would not fit the given info that 198/n² is an integer.)
So since n² > 0, 198/n² > 0
② How to start :
198/n² is a positive integer so n² must be a factor of 198.
We will need to break 198 into its prime factors first to find out which possible factors can n² be. Then we can find n.
198 ÷ 2 = 99
99 ÷ 3 = 33
33 ÷ 3 = 11 (prime)
So 198
= 11 × 3 × 3 × 2
= 11 × 3² × 2
These factors are the prime factors so they cannot be broken down/factorised into any smaller factors.
Now, n² is a positive integer so it must be the square of some other non-zero integer.
From above, we can see that 198 only has 3² being a prime factor with an even power.
This means that 11 and 2 cannot = n² since square rooting them would result in a non-integer value of n. We would get irrational numbers.
Eg.
√11 = 3.31662479....
√2 = 1.41421356...
These numbers are transcendental (the sequence of numbers never end)
3² = 9
9 is the only possible value of n² among them since 9 is a square number. This means that 9 can be square rooted to give an integer to satisfy the requirement for n.
198/n² = 198/9 = 22 →(a positive integer, fits the requirement)
And so ,
√3² = √9 = 3 is a possible value of n.
But,
Realise that 9 also = (-3) × (-3) = (-3)²
So if n² = 9, n could be 3 or -3.
i.e it is also possible that n = -√9 = -√3² = -3
So -3 is also a possible value of n.
① Some deductions first :
n and 198/n² are both integers.
This means that n² is also an integer since the product of two integers is always an integer.
n² is positive (n² > 0) regardless of whether n is negative or positive. Because , the square of any non-zero real value is always positive.
eg. 5² = 5 × 5 = 25
eg. (-10)² = (-10) × (-10) = 10 × 10 = 100
(Note that n cannot be 0 in this question. as that would mean n² = 0² = 0 .
Then this in turn implies that 198/n² = 198/0, which is undefined since division by 0 is undefined. This would not fit the given info that 198/n² is an integer.)
So since n² > 0, 198/n² > 0
② How to start :
198/n² is a positive integer so n² must be a factor of 198.
We will need to break 198 into its prime factors first to find out which possible factors can n² be. Then we can find n.
198 ÷ 2 = 99
99 ÷ 3 = 33
33 ÷ 3 = 11 (prime)
So 198
= 11 × 3 × 3 × 2
= 11 × 3² × 2
These factors are the prime factors so they cannot be broken down/factorised into any smaller factors.
Now, n² is a positive integer so it must be the square of some other non-zero integer.
From above, we can see that 198 only has 3² being a prime factor with an even power.
This means that 11 and 2 cannot = n² since square rooting them would result in a non-integer value of n. We would get irrational numbers.
Eg.
√11 = 3.31662479....
√2 = 1.41421356...
These numbers are transcendental (the sequence of numbers never end)
3² = 9
9 is the only possible value of n² among them since 9 is a square number. This means that 9 can be square rooted to give an integer to satisfy the requirement for n.
198/n² = 198/9 = 22 →(a positive integer, fits the requirement)
And so ,
√3² = √9 = 3 is a possible value of n.
But,
Realise that 9 also = (-3) × (-3) = (-3)²
So if n² = 9, n could be 3 or -3.
i.e it is also possible that n = -√9 = -√3² = -3
So -3 is also a possible value of n.
Now 198 also = 198 x 1
Then 198/1 = 198 →a positive integer
Since 1² = 1, then 1 is also a possible value of n² as it's a square number.
So possible value of n would be √1² = √1 = 1
But we also realise that (-1)² = (-1) × (-1)
= 1 × 1 = 1
So -1 is also a possible value of n.
Then 198/1 = 198 →a positive integer
Since 1² = 1, then 1 is also a possible value of n² as it's a square number.
So possible value of n would be √1² = √1 = 1
But we also realise that (-1)² = (-1) × (-1)
= 1 × 1 = 1
So -1 is also a possible value of n.
In short,
when n² = 9, n = √9 or n = -√9
(You will see this written as n = ±√9 sometimes
n = 3 or n = -3
When n² = 1,
n = √1 or n = -√1
n = 1 or n = -1
when n² = 9, n = √9 or n = -√9
(You will see this written as n = ±√9 sometimes
n = 3 or n = -3
When n² = 1,
n = √1 or n = -√1
n = 1 or n = -1
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