Subtracting radical expressions: 3^108-2^18-3^48 = ? (I am using the ^ for the square root symbol since my computer doesn't have one)
Hmm. could you perhaps draw out your question using the drawing tool below? :)
I'll take a stab at it. \[3\sqrt{108}-2\sqrt{18}-3\sqrt{48}\] Is that your question?/
Yes sorry I didn't know where the drawing tool is lol
Where is the drawing tool anyhow so I know for next time? Lol
You have your text book from which you are typing, and then 3 buttons below that: "equation" , "draw" and "attach"
Oh gotcha thank you.
So you have got:\[3\sqrt{108}-2\sqrt{18}-3\sqrt{48}\]108, 18, and 48 all have a common multiple of 6, so first tell me: 6 x what = 108? 6 x what =18? 6 x what = 48?
6x18=108 6x3=18 6x8=48
Alright
Ok but none of these seem to be a perfect square
6 x what = 18? 2 x what = 6? 2 x what = 8? 2 x what = 6?
Sorry my teacher isn't very good at explaining :(
6x3=18 2x3=6 2x4=8 2x3=6
We have to break it down by prime factorization. That is what we are doing now :)
Oh ok this seems much better thank you for taking the time to do this with me :)
Sorry my post was deleted -.- OS is glitching out for me.
No problem it happens lol
\[3\sqrt{108}-2\sqrt{18}-3\sqrt{48}\]\[3\sqrt{ 18 \cdot 6} -2\sqrt{6 \cdot 3} -3\sqrt{8 \cdot 6}\]\[3\sqrt{6\cdot\color{red}{ 3\cdot 3}}-2\sqrt{2\cdot \color{red}{3 \cdot 3}} - 3\sqrt{\color{red}{2 \cdot 2\cdot 2 \cdot 2}\cdot 3}\] Now you see all the highlighted pairs? For every 2 of these, 1 comes out of the square root.
When it's pulled out of the square root, it gets multiplied to the number in front of the square root. \[3 \cdot 3\sqrt{6} - 2 \cdot 3\sqrt{2} - 3\cdot 2 \cdot 2\sqrt{3}\] Now to get your answer you just simplify the numbers outside the parenthesis and you're done :)
Awesome thank you so much! You're great :)
No problem. Good luck!
Thanks :)
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