harmanvirdi90
Bunuel
Nityanshu1990
Wouldn't root 25 yeild +-5 as two values making it insufficient ?
When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.
Even roots have only a positive value on the GMAT. That is, \(\sqrt{25}=5\), NOT +5 or -5.
In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.
Mathematically
function: 2^x+(1/2^x) can never be negative for any value of x. So you can check this way. This is a more rational and appropriate approach rather than saying on GMAT "the only accepted answer is the positive root."
Not sure what you are trying to say there but what is written in my post is true and is an answer to the question asked by user "
is \(\sqrt{25}\) both 5 and -5".
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.