I just realized that
Find the value of |a|=a-3
If I go algebraically,
a>3 => no solution
0<a<3 => no solution because a>0 => a=a-3 => 0=(-3)
a<0 => not possible......(i)
However, if I square both the sides, a^2 = a^2 -6a + 9 => a = 3/2............(ii)
Why am I getting different answers for (i) and (ii)?
Can anyone please correct me what am I doing wrong?
Thanks
Voodoo
Modulus question (need expert help)
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Hi,
|a| = a-3.
As |a| is always >= 0, a>=3.
a can only take values greater than or equal to 3.
So, by squaring both sides, if you are getting a value for 'a'<3, then it should not be a solution. As a=3/2, which contradicts a>=3, there should be no solution.
I will give you an analogy
What is the solution for |x| = -2. We know |x| cannot be negative. So, straight away we can say no solution.
But squaring we get x^2 = 4 => x=-2 or +2.
Whenever we square both sides of equations involving modulus , first make sure the quantity equal to modulus is non-negative.
|a| = a-3.
As |a| is always >= 0, a>=3.
a can only take values greater than or equal to 3.
So, by squaring both sides, if you are getting a value for 'a'<3, then it should not be a solution. As a=3/2, which contradicts a>=3, there should be no solution.
I will give you an analogy
What is the solution for |x| = -2. We know |x| cannot be negative. So, straight away we can say no solution.
But squaring we get x^2 = 4 => x=-2 or +2.
Whenever we square both sides of equations involving modulus , first make sure the quantity equal to modulus is non-negative.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
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thanks for responding!
Let's take a = 4 ; |a-3| = 1 != 4. this equation doesnt have any solution. However, when you square it, it does have a solution!
I am not able to understand what's wrong.......
I dont agree with that. I thought about that already.a can only take values greater than or equal to 3.
Let's take a = 4 ; |a-3| = 1 != 4. this equation doesnt have any solution. However, when you square it, it does have a solution!
I am not able to understand what's wrong.......
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Hey,voodoo_child wrote:thanks for responding!
I dont agree with that. I thought about that already.a can only take values greater than or equal to 3.
I don't think you understood my point. Probably, my post wasn't too clear. I will try to elaborate it.
|a| = a-3 is given
LHS is >= 0 right?
So, RHS >= 0
So, a-3>=0. So, for finding the solution for this, you can only consider a>=3 as a<3 is definitely not going to give any solution.
So, for a<3 there is no solution
Now for a>=3, we solve this by your squaring method and we get a=3/2. But, this doesn't satisfy a>=3.
So, even for a>=3, we don't get any solution.
Why does |a-3| = 1 not have solution? Any typo?Let's take a = 4 ; |a-3| = 1 != 4. this equation doesnt have any solution. However, when you square it, it does have a solution!
I am not able to understand what's wrong.......
Anyway, this is not an illustration for the case we are dealing with. |a-3| = -1 would be an apt example.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
let me try another route
ques: |a|= a-3
For a>0 |a| = a
so a = a-3 or 0 = -3; No Solution
For a<0 |a| = -a
so -a = a-3 or 2a = 3 or a = 3/2
but a has to be negative, so there is no solution
As for squaring both sides, the basic condition that needs to be satisfied is that a-3 > 0 (because a-3 is equal to a mod function which is always positive), so a > 3
Solving through squaring gives a = 3/2 which is not >3 so no solution.
Hope this helps.
ques: |a|= a-3
For a>0 |a| = a
so a = a-3 or 0 = -3; No Solution
For a<0 |a| = -a
so -a = a-3 or 2a = 3 or a = 3/2
but a has to be negative, so there is no solution
As for squaring both sides, the basic condition that needs to be satisfied is that a-3 > 0 (because a-3 is equal to a mod function which is always positive), so a > 3
Solving through squaring gives a = 3/2 which is not >3 so no solution.
Hope this helps.