# Math Help - Quadratic Application

1.) The sum of a number and its reciprocal is $\frac{29}{10}$. Determine the integers.

So i figure, let $x$ be a number,
let $\frac{1}{x}$ be the reciprocal.

Now i got the equation:

$
\frac{x}{1}+\frac{1}{x}=\frac{29}{10}
$

From here i'm not sure what the LCD is in order to put it into a quad. equation. Then ill just be able to solve for x.

2.) 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.

What are "the zeros"? I dont know where to start with this one so if your in the mood to help explain im all ears

2. Originally Posted by Raj
1.) The sum of a number and its reciprocal is $\frac{29}{10}$. Determine the integers.

So i figure, let $x$ be a number,
let $\frac{1}{x}$ be the reciprocal.

Now i got the equation:

$
\frac{x}{1}+\frac{1}{x}=\frac{29}{10}
$

From here i'm not sure what the LCD is in order to put it into a quad. equation. Then ill just be able to solve for x.
Multily through by $x$ and rearrange:

$x^2-\frac{29}{10}x+1=0$

Solve this using the quadratic formula.

RonL

3. Originally Posted by Raj

2.) 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.

What are "the zeros"? I dont know where to start with this one so if your in the mood to help explain im all ears
These are the complex roots and so:

$
c/9= \frac{7 + \sqrt{-6}}{3}\ \times \ \frac{7 - \sqrt{-6}}{3}
$

and:

$
b/9 = \frac{7 + \sqrt{-6}}{3}\ + \ \frac{7 - \sqrt{-6}}{3}
$

RonL