Help neededThe quadratic $x^2+7x+20$ has the same roots as the quadratic $Ax^2+Bx+1$. What is $A+B$?

 2. p and q  are the solutions to the quadratic equation x^2 + 4x + 6 = 0. p^2 and q^2 are the solutions to the quadratic equation x^2 + bx + c = 0.Find b + c

pq=6

p+q=-4

\(p^2q^2=c\\ c=(pq)^2\\ c=6^2\\ c=36\\ p^2+q^2=-b\\ (p+q)^2=p^2+q^2+2pq\\ (-4)^2=-b+2*6\\ 16=12-b\\ b=-4\\~\\ b=-4 \;\;and \;\;c=36 \)

b+c = 32

1. The quadratic  x^2+7x+20 has the same roots as the quadratic Ax^2+Bx+1. What is A+B?

\(\mbox{For any quadratic equation } ax^2+bx+c\\ \mbox{The sum of the roots will be }\frac{-b}{a}\;\;and\\ \mbox{The product of the roots will be }\frac{c}{a}\)

So the sum ofthe roots will be  -7/1 = -7

and the product of the roots will be 20/1 = 20

\( \frac{1}{A}=20\\ A=\frac{1}{20}\\~\\ \frac{-B}{A}=-7\\ B \times \frac{1}{A}=7\\ 20B=7\\ B=\frac{7}{20}\\~\\ A=\frac{1}{20} \qquad B=\frac{7}{20} \)

1. The quadratic  x^2+7x+20 has the same roots as the quadratic Ax^2+Bx+1. What is A+B?

The roots of the first quadratic are   [-7 + i√31]  / 2        and  [-7 - i√31]  / 2

The product of these roots =    [49  + 31 ] /4  =  20 =  c / a

And in the second quadratic, c = 1....so    a = 1/20

And the sum of these roots  =  -7 =  -b/a

So  ....      -7   = -b / [ 1/20]      →   b = 7/20 

So.....a + b  =   1/20 + 7/20  =    8 / 20    =  2 / 5  

 2. p and q  are the solutions to the quadratic equation x^2 + 4x + 6 = 0. p^2 and q^2 are the solutions to the quadratic equation x^2 + bx + c = 0.Find b + c

The solutions to    x^2 + 4x + 6  = 0    are    -2 + i √2     and   -2 - i √2

Let the first be p and the second one, q

So  p^2   =  2 - 4i √2      and q^2   = 2 + 4i √2

The product of these roots =  36  =  c / a

And a   = 1, so c = 36

And the sum of these roots =  4  = -b / a

And a = 1, so b = -4

So....b + c  =  -4 + 36   =  32