+128448 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
+1253 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
