farukqmul wrote:
Bunuel wrote:
What does |2b| equal?
(1) b^2-|b|-20=0. Solve quadratics for \(|b|\): \((|b|)^2-|b|-20=0\) --> \(|b|=-4\) or \(|b|=5\). Since absolute value cannot be negative then we have that \(|b|=5\) and \(|2b|=10\). Sufficient.
(2) |2b|=3b+25. Two cases:
If \(b\leq{0}\) then we would have that \(-2b=3b+25\) --> \(b=-5\).
If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.
So, we have that \(b=-5\), hence \(|2b|=10\). Sufficient.
Answer: D.
Hope it's clear.
Hi, can you make the solution more clear?in case of (1) why you didn't consider the negative value of b? When b is positive, b^2 -b = 20 and when b is negative b^2 + b =20..then we will find 4 values of b...I am confused ..Can you explain? and in case of (2) how you discard the 2nd value? Thanks
OK.
Say \(x=|b|\), then we have that \(x^2-x-20=0\) --> \(x=-4\) or \(x=5\). Now, \(x=|b|=-4\) is not possible, since an absolute value of a number (\(|b|\)) cannot be negative. So, we have that \(|b|=5\) and \(|2b|=10\).
Now, you can solve this statement considering two ranges: \(b\leq{0}\) and \(b>0\), which will lead you to the same.
If \(b\leq{0}\) then we'll have \(b^2+b-20=0\) --> \(b=-5\) or \(b=4\) (not a valid solution since we are considering the range when \(b\leq{0}\));
If \(b>{0}\) then we'll have \(b^2-b-20=0\) --> \(b=-4\) (not a valid solution since we are considering the range when \(b>{0}\)) or \(b=5\);
So, only two valid solutions: \(b=-5\) or \(b=5\) --> \(|2b|=10\).
As for (2), it's explained in the post:
If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but
since we are considering the rangeo when \(b>{0}\) then discard this solution.
Hope it's clear.
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