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# Help

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

Jun 15, 2019

#1
+106533
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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

Jun 15, 2019
#2
+999
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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