thegame12 wrote:
Hello everyone,
i don't get it how we can't switch from the 3rd equation to the last one. √b²=b, no ?
Interesting question. First of all, I'd say that you
can switch from the third equation to the last one. You'd use FOIL (or a special quadratic that you have memorized). The earlier posters showed how that works.
If you want to double check that on your own, try plugging some numbers into both equations. If you get different results from the two equations, then you'd know that they weren't actually the same. But if you always get the same result from both of them, that would make you think that they're probably the same.
For example, try plugging in b = 0.
\(a = (1-\sqrt{0})^2 = 1^2 = 1\)
\(a = 1-2\sqrt{0}+0 = 1-2(0)+0 = 1\)
Okay, it's the same in that case. How about b = 1?
\(a = (1-\sqrt{1})^2 = 0^2 = 0\)
\(a = 1-2\sqrt{1}+1 = 1-2(1)+1 = 0\)
It's the same in that case too. What if b = 100?
\(a = (1-\sqrt{100})^2 = (-9)^2 = 81\)
\(a = 1-2\sqrt{100}+100 = 1-2(10)+100 = 81\)
It's still the same.
There's something slightly odd going on here though, which is what I think you may be asking about. Notice that I'm not plugging in any negative values for b. If I did that, I would have a problem. But the problem isn't that the two equations are different. It's just that (on the GMAT) you can't take the square root of a negative number at all. So, we don't really need to worry about negative numbers here. The GMAT will never give you expressions like these ones and ask you to use negative values with them.
Another, similar issue - but
not the issue on this problem - happens when you do \(\sqrt{x^2}\).
The GMAT has a rule that says that we only look at positive square roots. So, \(\sqrt{100}\) is just 10, not '10 or -10', like you may have learned in school. Since we don't know whether \(x\) is positive or negative, we can't just say that \(\sqrt{x^2} = x\). (The problem is that \(x\) might be a negative number, which would make the equation wrong.) Instead, you have to take the absolute value of x to make sure it comes out positive: \(\sqrt{x^2} = |x|\)
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