VeritasPrepKarishma wrote:
dave13 wrote:
regarding this |x - 3| well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3?
\(x\) could be -5 no
?
can you advice in which case can i square both sides od modulus ? can i do it |x+4| + |x - 3| = 10
How about this question \(|x-2| + |x+5| - |x-3| > |x+13|\) woukd i be able to square both sides ? if not please explain why ?
thank you in advance for your kind explanation and have an awesome day
Hey Dave,
I am not sure I understand your question but here is something that might help you:
Why do we take the range as 'x - 3 > 0' ? or 'x + 5 > 0' ?
The definition of absolute value will help you understand this. It is discussed here:
https://www.veritasprep.com/blog/2014/0 ... -the-gmat/Can you square |x+4| + |x - 3| = 10?
Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as |x+4| + |x - 3| > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign.
\(|x+4| + |x - 3| = 10\)
\(|x + 4|^2 + |x - 3|^2 + 2*|x+4|*|x - 3| = 10^2\)
The term 2*|x+4|*|x - 3| will continue to have absolute value sign. How will you take care of it?
Hi
VeritasPrepKarishma, many thanks for you reply and the article
i wish i could answer your question,
i dont know how i would take care off it but there is one math wizzard called
pushpitkc (perhaps he learnt from you
) who actually applied techique of squaring modulus in this question
https://gmatclub.com/forum/how-many-val ... l#p2011543 i extracted his solution here (in wine red
with question stem and answer choices ), so you dont overload your laptop with many internet browser tabs opened
otherwise your laptop risks to be frozen like mine
so see below this magician squares modulus
so my question when can i apply squaring to modulus ? in which cases can I apply and which cases it wont work ? How many values of x satisfy the equation ||x-7|-9|=11?
A. 1
B. 2
C. 3
D. 4
E. 5
If |x-7| = a, the equation will become |a - 9| = 11 ( how did he break this equation into two btw ?
Squaring on both sides,
\(a^2\)−18a+81=121
\(a^2\)−18a+81=121
\(a^2\)−18a−40=0
\(a^2\)−18a−40=0
Solving for a, a = 20 or -2
If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.
Therefore, there are 2 values of x which satisfy the equation(Option B) Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign.
\(|x|^2 = (x)^2 = (-x)^2\)
\(|x + 2|^2 = (x + 2)^2 = x^2 + 2^2 + 2*x*2\) (No absolute value sign left)
Look at the example you gave:
\(||x-7|-9|=11\)
Just denote |x - 7| as a. You get
\(|a - 9| = 11\)
Square it and get 'a'.
But you don't need a. You need the value of x. For that, we know
\(a = 20 = |x - 7|\) (as assumed before)
Squaring again will get the answer.
How about an expression such as
\(|2x + 3| = |x - 4|\)
Again squaring will help since absolute value sign from both sides will go away.
How about \(|x+4| + |x - 3| = 10\) When you square the left hand side, you use \((a + b)^2 = a^2 + b^2 + 2ab\) where \(a = |x + 4|\) and \(b = |x - 3|\)
So \((|x+4| + |x - 3| )^2 = |x + 4|^2 + |x - 3|^2 + 2*|x + 4|*|x - 3|\)
\(= (x + 4)^2 + (x - 3)^2 + 2*|x + 4|*|x - 3|\)
Here is the problem. 2*|x + 4|*|x - 3| still has absolute value signs.
_________________
Karishma
Veritas Prep GMAT Instructor
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