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December 22, 2014

December 22, 2014

Posted by **Tony** on Thursday, May 23, 2013 at 3:39pm.

- Maths -
**Steve**, Thursday, May 23, 2013 at 3:49pm19x^2+3x+2=0 has roots

(-3±√143 i)/38

since both roots are complex, if the two quadratics share one root, they share both. So,

Ax^2+BX+7 must be a multiple of 19x^2+3x+2.

So, it must be 7/2 times, making it

19(7/2)x^2 + 3(7/2)x + 2(7/2)

= 66.5x^2 + 10.5x + 7

A+B = 77

- Maths -
**Tony**, Thursday, May 23, 2013 at 3:51pmthank you very much

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