If you’re on the East Coast, I hope that you and yours are able to weather the storm OK. I’m not sure whether I’ll have power or not on Monday, but I’ll try to get the solution posted then if possible.

This week’s prize will again be access to the Math Guide Beta, which is looking radder and radder by the day. I’ve been doing a TON of work on it. Anyhoo, first correct (and *not anonymous*) answer in the comments gets access.

If (

x^{2}+ 5x)^{2}– 36 = (x+m)(x+n)(x+p)(x+q), what is the median of the set {m,n,p,q}?

Good luck, kiddos. And stay safe.

UPDATE: Nice work, Ammad. I hope you enjoy the book, which I’ve now shared with you in Google Docs.

Solution below the cut.

The trick to getting this one right (or at least, to getting it right without the aid of a powerful calculator) is recognizing that the left side of the original equation *is actually the difference of two squares*:

*x*

^{2}+ 5

*x*)

^{2}– 36

*x*

^{2}+ 5

*x*+ 6)(

*x*

^{2}+ 5

*x*– 6)

*x*+ 3)(

*x*+ 2)(

*x*+ 6)(

*x*– 1)

*do*know that {m, n, p, q} = {-1, 2, 3, 6}, which has a median of 2.5.

*two*cookies.

## Comments (6)

(x-1)(x+2)(x+3)(x+6)

m = (-1), n = 2, p = 3, q =6

median = 2.5

Excellent work! Welcome to the Beta!

Thanks!

mnpq = 2x2x3x3 = 36

median =(2+3)/2

median = 2.5

I know this is a bit late but i believe the median could also be 2.

I plugged in 2 for x and got 160 on the left side and then played around with numbers for m n p and q until I got m=0 n=2 p=2 and q=3

Here’s my equation to prove it

160=(2+0)(2+2)(2+2)(2+3)

160=(2)(4)(4)(5)

160=160

And the median of 0,2,2 and 3 would be 2

Would this be another answer or did i just mess up somewhere

Well, I suppose that could work for that one value of x, but those same numbers won’t work for other values of x. I wrote this question over 3 years ago, and admittedly I wasn’t careful enough in my wording. It should definitely say the the equation is true for all values of x. :/