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[quote=bobbym]Hi; [hide=answer] Turn it into a monic polynomial: [math]x^2+\frac{21 x}{18}-\frac{400}{18}=0[/math] If the roots are a and b then we can say [math](x -a)(x-b)=0[/math] [math](x -a)(x-b)= x^2 -(a+b)x+ab[/math] equating coefficients [math] (a+b) = \frac{-21}{18}[/math] [math]ab = \frac{-400}{18}[/math] Now we know that [math](a + b)^2 = a^2 + 2ab + b^2[/math] [math](a + b)^2 - 2ab = a^2 + b^2[/math] just plugging in [math]\left(-\frac{21}{18}\right)^2-\frac{2 (-400)}{18}= \frac{1649}{36}[/math] So the sum of the squares of the roots is 1649 / 36 [/hide][/quote]
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bobbym
2013-02-01 19:11:56
Hi;
cooljackiec
2013-02-01 14:57:23
What is the sum of the squares of the roots of
?
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