jcallahan6 wrote:
Hello,
I am having trouble understanding why when factoring in a root you would leave the power to a number in the original number you factored.
Ex. \sqrt{36^2+15^2}. When factoring this the answer noted to keep the power of the factored number. --> \sqrt{3^2(12^2+5^2)}. Can someone please explain to me why it wouldn't look like this --> \sqrt{3^2(12+5)}. Is there a rule I am missing? I noticed that when it involves factoring an expression in a root that it keeps the power, but if it isn't in a root it doesn't. Please correct me if I a wrong here. Please provide a detailed explanation.
Original question: \sqrt{36^2+15^2} simply the root and express answer as an integer.
Best,
Jai
Hi!
I'm not sure what you exactly mean, but let's go step by step.
\(\sqrt{(36^2+15^2)}\)=\(\sqrt{((3*3*2*2)^2+(3*5)^2)}\)=\(\sqrt{((3^2*2^2)^2+(3*5)^2)}\)=\(\sqrt{3^4*2^4+3^2*5^2}\)=\(\sqrt{3^2(3^2*2^4+5^2)}\)=
\(\sqrt{3^2((3*2^2)^2+5^2)}\)=\(\sqrt{3^2(12^2+5^2)}\)=\(\sqrt{3^2(144+25)}\)=\(\sqrt{3^2(169)}\)=\(\sqrt{3^2*13^2}\)=3*13=69
if you forget how to multiply, use simple numbers: \(3^2*3^2=3*3*3*3=3^4\). So, basically, you have to add power to power - 2+2.
Thank you for the help! I see how you broke it down into primes to better explain this. The book answer left me a bit confused so it was nice to see it this way as as a better way of understanding to use simple numbers.