Solving Equations with Roots

On Thursday we were finished with our self-guided projects on 5-9 and 5-10 and moved on the more math. Thursday also happened to be Pi day so we had some very good cookies and had some Pi related fun. We practiced reciting the digits from Pi, everyone did fairly well, and two students reached of 40 digits!

When we were done with our fun we moved on to learning about solving equations that had roots in them.

This little symbol is the square root sign. The solution then becomes whatever will square to become x. For example, the square root of 36 is 6, because 6^2 = 36.

Here is another image which displays the “index” part of roots. Whatever is in the index is the exponent that the solution would have to equal the radicand. In the example above, the solution is 3, because 3^3 = 27.

Here is an example of a problem where we simplified the square root of 32. 32 is not a perfect square, so we had to find a perfect square that was a factor of 32. 2 and 16 multiply to 32 so we can separate the two and then find the square root of 16, which is 4. So that means our simplified expression is 4*sqrt2.

In these four problems all you need to do is simplify within the root. In the third problem, we see -1^2 become 1, this is because when two negatives multiply, they become positive. When you have one term by itself that is squared, you can cancel it, but not if there are multiple terms. In problem one you can’t cancel because there is subtraction but in problem 4 you can cancel the square.

When you have fractions inside a root symbol, you can simplify the numerator and the denominator. We can see in problem 2 an example of solving with root 3. 3*3 equals 9, and 9*3 equals 27 so 27 becomes 3. 5*5 equals 25 and 25*5 equals 125 so 125 becomes 5. The simplified expression is 3/5.

Thanks for reading my blog! I hope you learned a lot!