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What is the sum of the squares of the roots of $18x^2 +21x - 400$?

 Jun 15, 2019
 #1
avatar+104962 
+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

 

 

cool cool cool

 Jun 15, 2019
 #2
avatar+974 
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

 Jun 15, 2019

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