#1**+1 **

18x^2 + 21x - 400

Let r and s be the roots

The sum of the roots , r + s , = -21/ 8

The product of the roots, rs = -400/18 = -200/9

So

(r + s)^2 = (-21/8)^2

r^2 + 2rs + s^2 = (441/64)

r^2 + 2(-200/9) + s^2 = 441/64

r^2 + s^2 - 400/9 = 441/64

r^2 + s^2 = 441/64 + 400/9

r^2 + s^2 = 29569 / 576

CPhill Jun 15, 2019

#2**0 **

First thing is to find the roots.

We can use the quadratic formula to find the roots

\(\frac{-21\pm\sqrt{21^2-4(18)(-400)}}{2\times 18} = \frac{-21\pm 171}{36}\)

x equals 25/6 or -16/3.

\((\frac{25}{6})^2+(-\frac{16}{3})^2=\frac{625}{36}+\frac{256}{9}=\frac{1649}{36}\)

You are very welcome!

:P

CoolStuffYT Jun 15, 2019