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

gueesstt  Apr 9, 2018
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 #1
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let the roots be x and y. we know that x+y=-b/a and xy=c/a. the sum of the squares of the roots is x2 + y2 and we want to rearrange this to be a comination of (x+y) and xy. (x+y)2 = x2 + y2 + 2xy. therefore, we can write x2 + y2 as (x+y)2 -2xy. subtitute in the other answers from the sum of the roots and the product of the roots gives you (-b/a)2 -2(c/a). in the quadratic 18x2 +21x-400, a=18,b=21,c=400. substitute those values in and you get (-21/18)2 -2(-400/18). hope that helps.

nadja.birch1  Apr 9, 2018
 #2
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Solve for x:
18 x^2 + 21 x - 400 = 0

 

The left hand side factors into a product with two terms:
(3 x + 16) (6 x - 25) = 0

 

Split into two equations:
3 x + 16 = 0 or 6 x - 25 = 0

 

Subtract 16 from both sides:
3 x = -16 or 6 x - 25 = 0

 

Divide both sides by 3:
x = -16/3 or 6 x - 25 = 0

 

Add 25 to both sides:
x = -16/3 or 6 x = 25

 

Divide both sides by 6:

x = -16/3    or    x = 25/6     Sum of the squares =[-16/3]^2 + [25/6]^2 =1649 / 36

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