Question 346670
I don't think you are supposed to use a computer or calculator, even if you remember how to use them to find square roots. All computers or calculators can do with most square roots is to find a decimal approximation for the square root.<br>
Instead, I believe your problem is to simplify the expression. Simplifying square roots involves finding perfect square factors of the radicand, if any. ("Radicand" is the name for the expression inside a radical (like a square root).) So we will start by looking for perfect square factors in each of the radicands and, if we find them, rewriting the radicands as a product involving these perfect squares:
{{{sqrt(150) - sqrt(24) -sqrt(54)}}}
{{{sqrt(25*6) - sqrt(4*6) - sqrt(9*6)}}}
Next we use a property of radicals, {{{sqrt(a*b) = sqrt(a)*sqrt(b)}}}, to separate the perfect squares factors into their own square roots:
{{{sqrt(25)*sqrt(6) - sqrt(4)*sqrt(6) - sqrt(9)*sqrt(6)}}}
Now we can replace the square roots of the perfect squares:
{{{5sqrt(6) - 2sqrt(6) - 3sqrt(6)}}}
Now the square roots have been simplified. Next we simplify the expression as a whole, if possible. It turns out that these are like terms so they can be added/subtracted. (If you have trouble seeing this, perhaps using a temporary variable will help. Let {{{q = sqrt(6)}}}. Then the expression becomes: {{{5q - 2q -3q}}}. This, I hope you can see, is an expression of like terms. And so is the expression with the square roots of 6!)<br>
As it turns out, subtracting {{{2sqrt(6)}}} from {{{5sqrt(6)}}} leaves {{{3sqrt(6)}}} (just like 5q - 2q  = 3q). And if we then subtract {{{3sqrt(6)}}} from {{{3sqrt(6)}}} we get 0.<br>
So the simplified expression is zero!