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- Feb 14, 2012

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Compute $\sqrt{2000(2007)(2008)(2015)+784}$ without the help of calculator.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,906

Compute $\sqrt{2000(2007)(2008)(2015)+784}$ without the help of calculator.

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\(\displaystyle 2007\cdot2008=4030028+28\)

\(\displaystyle 784=28^2\)

Hence:

\(\displaystyle 2000\cdot2007\cdot2008\cdot2015+784=4030028^2-28^2+28^2=4030028^2\)

And so:

\(\displaystyle \sqrt{2000\cdot2007\cdot2008\cdot2015+784}=\sqrt{4030028^2}=4030028\)

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- #3

- Feb 14, 2012

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Bravo,

- Mar 31, 2013

- 1,346

we have 2000 * 2007 *2008 * 2015 + 784

a(a+7)(a+8)(a+15) + 784

= a(a+15) (a+7) (a+8) + 784

= (a^2+15a) (a^2 + 15a + 56) + 784

= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$

= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$

= $(a^2 + 15 a + 28)^2$

hence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28

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- #5

- Feb 14, 2012

- 3,906

Hey

we have 2000 * 2007 *2008 * 2015 + 784

a(a+7)(a+8)(a+15) + 784

= a(a+15) (a+7) (a+8) + 784

= (a^2+15a) (a^2 + 15a + 56) + 784

= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$

= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$

= $(a^2 + 15 a + 28)^2$

hence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28

- Feb 9, 2012

- 33

$ \sqrt{(2000)(2007)(2008)(2015)+ 784} \ = $

$ \sqrt{16 [ (\frac{2000}{4})(2007) ( \frac{2008}{4} ) (2015) + \frac{16 \cdot 7^2}{16} ]} \ = $

$ 4 \sqrt{(500)(2015)(502)(2007) \ + \ 7^2} \ = $

$ 4 \sqrt{(1007507 -7)(1007507 + 7) + 7^2} \ = $

$ 4 \sqrt{(1007507)^2} \ = \ 4030028 $

Admittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.

Last edited:

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- #7

- Feb 14, 2012

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Nice one,

$ \sqrt{(2000)(2007)(2008)(2015)+ 784} \ = $

$ \sqrt{16 [ (\frac{2000}{4})(2007) ( \frac{2008}{4} ) (2015) + \frac{16 \cdot 7^2}{16} ]} \ = $

$ 4 \sqrt{(500)(2015)(502)(2007) \ + \ 7^2} \ = $

$ 4 \sqrt{(1007507 -7)(1007507 + 7) + 7^2} \ = $

$ 4 \sqrt{(1007507)^2} \ = \ 4030028 $

Admittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.