parthAgarwal3033
I still don't get why Root(20) can't be negative. (-4.47)^2 is also equal to 20. Also if we solve x^2 = 20 we get 2 real values of x (4.47 and -4.47) Now if we consider integers between -4.47 and 5.33 , there can be many values right??
Mathematically, \(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.
The graph of the function f(x) = √xNotice that it's defined for non-negative numbers and is producing non-negative results.
TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:
\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
Hope it helps.