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Algebra Level 2

$\large \sqrt{(2000\times 2008\times 2009\times 2017)+ 1296} = \ ?$

3 solutions

Mohsin Naqvi
Jun 24, 2015

I replaced 2000 with $x$ . This makes $2000 \times 2008 \times 2009 \times 2009$ = $x \times (x+8) \times (x+9) \times (x+17)$ .

I simplified it as $(x^{2} +17x) \times (x^{2}+17x+72)$

further replacing $x^{2}+17x$ with $y$ , I get $y \times (y+72)$

putting it back in the original problem, we get $\sqrt{y^{2}+72y+1296}$

Solving we get the answer $y+36$

since $y = x^{2}+17x$ , we get $x^{2}+17x+36$

since $x = 2000$ , we get the answer as 4034036

Abhisek Mohanty
Jun 1, 2015

Ya please send a solution to know how to do it ......................

I have posted a solution.

Jonathan Hsu - 6 years ago

Jonathan Hsu
May 31, 2015

Ashwin- write a solution to this problem, so I know you solved it legit

Can you write out the solution yourself? It would certainly benefit the community!

We have: $\large \sqrt{(2000\times 2008\times 2009\times 2017)+ 1296} = \ ?$

Set 2000 as x: $\large \sqrt{(x\times (x+8) \times (x+9)\times (x+17))+ 1296} = \ ?$

Now:

$\large \sqrt{(((x+8)\times (x+9))\times ((x+17)\times (x)))+ 1296} = \ ?$

= $\large \sqrt{(x^2 + 17x + 72)\times (x^2 + 17x)+ 1296} = \ ?$

= $\large \sqrt{(x^2 + 17x + 36 + 36)\times (x^2 + 17x + 36 - 36)+ 1296} = \ ?$

Now we can set $\large (x^2 + 17x + 36)$ as A and $\large 36$ as B.

$\large((A+B)(A-B) = A^2 - B^2)$

= $\large \sqrt{(x^2 + 17x + 36)^2 - 36^2 + 36^2)}$

= $\large (x^2 + 17x + 36)^2$

Plugging 2000 for x gives us 4034036.

Jonathan Hsu - 6 years ago

In step 2, your order of operations is wrong

Ashwin Padaki - 6 years ago

thanks, I will fix it

Jonathan Hsu - 6 years ago

I did solve it legit.

Ashwin Padaki - 6 years ago