Question

Divide 8 into 2 parts such that 7 times the square of the smaller part exceeds twoce the square of the other part by 13

Answer 1

a + b = 8

7 a^2 = 2 b^2 + 13

7 (8 - b)^2 = 2 b^2 + 13

448 - 112 b + 7 b^2 = 2 b^2 + 13

5 b^2 - 112 b + 435 = 0

Answer 2

smaller part --- x
larger part ---- 8 - x
7x^2 > 2(8-x)^2 by 13
7x^2 = 2(8-x)^2 + 13
7x^2 = 128 - 32x + 2x^2 + 13
5x^2 + 32x - 141 = 0
(x - 3)(5x + 47) = 0
x = 3 or x = -47/5

the two parts are either 3 and 5
or
they are -47/5 and 87/5