Quadratic Application

**Raj** · October 12th 2007, 01:03 PM

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

**CaptainBlack** · October 12th 2007, 01:15 PM

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

**CaptainBlack** · October 12th 2007, 01:19 PM

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

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