threepointonefourone

Using brute force , which is something you should NOT do (even though this time, I'm just lazy):

\(11250x^6 + 12750(x^5)(y) - 1800(x^4)(y^2) - 4740(xy)^3\)\( - 1662(x^2)(y^4)-234(x)(y^5) - 12y^6\)

\(\text{The coefficient of} (x^2)(y^4) \text{​ is -1,662.}\)

Hope this helped .

2 = 2

2 + 7 = 9

2 + 7 + 12 = 21

2 + 7 + 12 + 17 = 38

2 + 7 + 12 + 17 + 22 = 60

2 + 7 + 12 + 17 + 22 + 27 = 87

2 + 7 + 12 + 17 + 22 + 27 + 32 = 119

The answer is 32.

Let a1 be the first term

Let d be the difference between each term (that is, a2 - a1, or a3 - a2, etc)

a1 + (a1 + 2d) = -2

(a1 + d) + (a1 + 3d) + (a1 + 5d) = 1

Simplify first equation:

2a1 + 2d = -2

a1 + d = -2

Wait what? we see a1 + d = a2 = -2 !!!

Simplify second equation

-2 + (2a1 + 8d) = 1

2a1 + 8d = 3

minus 2(a1 + d) = 2(-2)

6d = 3 - (-4)

d = 6/7

since a2 = -2, 

d = -2 - 6/7 = -2 6/7 = -20/7

Isn't it 100~

this is close to 1 *10^100, BUT it's actually 9.9999... * 10^99

\(\text{By Vieta's formulae the sum of roots is -b/a, } \text{and the product is c/a}\)

if they are in an quadratic ax^2 + bx + c = 0

WE KNOW A = 1 :)

Therefore, the sum of the roots is -b and the product is c

The roots are 1 and -1, so the sum is 0 and the product is -1

Given that -b = 0 and c = -1, or b = 0 and c = 1, plug the values back into the equation x^2 + bx + c = 0 to get your final answer

Please like if this answer was correct. Thank you.

\(\text{By graphing this on Desmos, you get -1.119 or 4.619, and we choose the bigger one, or 4.619} \)