19x^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
A and B are real numbers such that the two quadratic equations 19x^2+3x+2=0 and Ax^2+Bx+7=0 have a common root. What is the value of A+B
2 answers
thank you very much