Algebra 2

The roots of $x^2+8x+4$ are the same as the roots of $Ax^2+Bx+1$. What is $A+B$?

I suggest you enter questions 2 and 3 as separate questions.

1.

The roots of $x^2+8x+4$ are the same as the roots of $Ax^2+Bx+1$. What is $A+B$?

The roots:

$\begin{array}{|rcll|} \hline \mathbf{x^2+8x+4} &=& \mathbf{0} \\\\ x &=& \dfrac{-8 \pm \sqrt{8^2-4*4} }{2} \\ \\ x &=& \dfrac{-8 \pm \sqrt{48} }{2} \\ \\ x &=& \dfrac{-8 \pm \sqrt{3*16} }{2} \\ \\ x &=& \dfrac{-8 \pm 4\sqrt{3} }{2} \\ \\ \mathbf{x_1} &=& \mathbf{-4 + 2\sqrt{3}} \\ \mathbf{x_2} &=& \mathbf{-4 - 2\sqrt{3}} \\\\ \mathbf{x_1x_2} &=& (2\sqrt{3}-4)\left( (-1)(2\sqrt{3}+4) \right) \\ &=& -(2\sqrt{3}-4) (2\sqrt{3}+4) \\ &=& -\left(4*3-16 \right) \\ &=& -\left(-4 \right) \\ \mathbf{x_1x_2} &=& \mathbf{4} \\\\ \mathbf{x_1+x_2} &=& -4 + 2\sqrt{3} -4 - 2\sqrt{3} \\ &=& -4 -4 \\ \mathbf{x_1+x_2} &=& \mathbf{-8} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{ Ax^2+Bx+1 } &=& \mathbf{ 0 } \quad | \quad : A \\\\ x^2+ \underbrace{\dfrac{B}{A}}_{=-(x_1+x_2)}x+ \underbrace{\dfrac{1}{A}}_{=x_1x_2} &=& 0 \\\\ \hline \mathbf{ \dfrac{1}{A} } &=& \mathbf{x_1x_2} \quad | \quad \mathbf{x_1x_2=4} \\\\ \dfrac{1}{A} &=& 4 \\\\ \mathbf{ A } &=& \mathbf{\dfrac{1}{4}} \\ \hline \mathbf{\dfrac{B}{A}} &=& -(x_1+x_2) \quad | \quad \mathbf{x_1+x_2=-8} \\\\ \dfrac{B}{A} &=& -(-8) \\\\ B &=& 8A \\ B &=& 8*\dfrac{1}{4} \\ \mathbf{ B} &=& \mathbf{2} \\ \hline A+B &=& \dfrac{1}{4} + 2 \\ \mathbf{A+B} &=& \mathbf{2.25} \\ \hline \end{array}$