1. Quad. Eq. Need a check+help

Determine the roots of the following. Simplified radical form.

A) $\frac{x-1}{x+4} - \frac{x}{x-3}=9$

$(x-1)(x-3) - (x)(x+4) = 9(x+4)(x-3)$
$x^2-4x+3 - x^2+4x = 9x^2+9x-108$
$-8x+3 = 9x^2=9x^2+9x-188$
$9x^2+17x-111=0$

Then i used the quad formula. My only question is, in the above second line do i distribute the - to the +4x? Or do i keep it positive? I went and made it -4x to get the below answer.

$x = \frac{-17 \pm \sqrt{4285}}{18}
$

and then if its not any trouble

7. The quadratic function $y=9x^2+bx+c$ has no x-intercepts but the zeros of the function are $\frac{7 \pm \sqrt{-6}}{3}$. Determine the values of b and c. Show all work.

I don't know whats going on here so explain if you feel like it

2. Hello, Raj!

Determine the roots of the following.

$A)\;\;\frac{x-1}{x+4} - \frac{x}{x-3}=9$

$(x-1)(x-3) - x(x+4) \:= \:9(x+4)(x-3)$

$x^2-4x+3 - x^2+4x \:= \:9x^2+9x-108$

$9x^2+17x-111=0$

Then i used the quad formula: . $x = \frac{-17 \pm \sqrt{4285}}{18}$
You did great!

7. The quadratic function $y\:=\:9x^2+bx+c$ has zeros: . $\frac{7 \pm \sqrt{\text{-}6}}{3}$.
Determine the values of b and c.
We can run the problem "backwards" . . .

The roots are: . $\begin{array}{ccc}x & = & \frac{7 + \sqrt{\text{-}6}}{3} \\ x & = & \frac{7 - \sqrt{\text{-}6}}{3}\end{array}$

This gives us: . $\begin{array}{ccccccc}x - \frac{7 + \sqrt{\text{-}6}}{3} \: = \: 0 & \Rightarrow & 3x - 7 + \sqrt{\text{-}6} \: = \: 0 \\ x - \frac{7-\sqrt{\text{-}6}}{3} \:= \:0 & \Rightarrow & 3x - 7 - \sqrt{\text{-}6} \:= \:0\end{array}$

Multiply the two equations: . $\left[(3x-7) + \sqrt{\text{-}6}\right]\left[(3x-7) - \sqrt{\text{-}6}\right] \:=\:0$

. . and we have: . $(3x-7)^2 - (\sqrt{\text{-}6})^2 \:=\:0\quad\Rightarrow\quad 9x^2 - 42x + 49 - (\text{-}6) \:=\:0$

Hence, the function is: . $y \;=\;9x^2 - 42x + 55$

. . Therefore: . $\boxed{b \,= \,-42,\;c \,= \,55}$