What is the fastest way to solve sqroot of (16)(20)+(8)(32)?
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…other than sq root of (320)+(256) = sq root of 576 = 24
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Your method looks fast enough, but for bigger numbers (or if you are not comfortable with square roots) you may want to try something like this:
sqrt[(16.32+8.32)]=sqrt[(4.4.2.2.5+4.4.2.2.4)]=sqrt[4.4.2.2.(5+4)]=
sqrt[(4.4.2.2.3.3)]= 4.2.3=24
sqrt[(16.32+8.32)]=sqrt[(4.4.2.2.5+4.4.2.2.4)]=sqrt[4.4.2.2.(5+4)]=
sqrt[(4.4.2.2.3.3)]= 4.2.3=24
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16x20 + 8x32 ( you should see that you can take x2 from 32 and give it to 8)
16x20 + 16x16
16x (20+16)
16x36
So you can take these two numbers out of root as:
4x6 = 24
16x20 + 16x16
16x (20+16)
16x36
So you can take these two numbers out of root as:
4x6 = 24
LGTCH
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on the same line as logitech, when it comes to big numbers, and since I don't memorize the square root of large numbers, I use a combination of muzali and logitech's methods.
for this example, I'd break out your question to:
sqrt[(4x4x4x5+4x4x4x4)]=sqrt[(4x4x4x(5+4)] (I factored out 4x4x4) =sqrt[(4.4.4.(9)]
I have a simple mind, so when i see sqrt[(4x4x4x9], I can easily figure out the square root of each. the answer is therefore, 2x2x2x3=24
for this example, I'd break out your question to:
sqrt[(4x4x4x5+4x4x4x4)]=sqrt[(4x4x4x(5+4)] (I factored out 4x4x4) =sqrt[(4.4.4.(9)]
I have a simple mind, so when i see sqrt[(4x4x4x9], I can easily figure out the square root of each. the answer is therefore, 2x2x2x3=24