Help with daughter's homework

**Embalmed** · February 5th 2008, 07:52 PM

I was trying to help my daughter solve a problem from her algebra homework. The answer I found through trial and error as well as from another site (c = 40.5511), but the steps to get there have confounded me.

$0.01 = -0.027c^2 + 8c - 280$

Any guidance or suggestions would be appreciated.

**Zach** · February 5th 2008, 08:46 PM

Try wrapping [ math] and [/ math] tags around the problem, it's easier to read :P

(remove spaces in tags)

**Jhevon** · February 5th 2008, 08:58 PM

Originally Posted by Embalmed

I was trying to help my daughter solve a problem from her algebra homework. The answer I found through trial and error as well as from another site (c = 40.5511), but the steps to get there have confounded me.

$0.01 = -0.027c^2 + 8c - 280$

Any guidance or suggestions would be appreciated.

first subtract 0.01 from both sides to get $-0.027c^2 + 8c - 280.01 = 0$

now use the quadratic formula

the quadratic formula says, the solutions to a quadratic equation of the form $ax^2 + bx + c = 0$ are given by:

$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

can you continue?

**earboth** · February 5th 2008, 09:00 PM

Originally Posted by Embalmed

I was trying to help my daughter solve a problem from her algebra homework. The answer I found through trial and error as well as from another site (c = 40.5511), but the steps to get there have confounded me.

$0.01 = -0.027c^2 + 8c - 280$

Any guidance or suggestions would be appreciated.

Hi,

this is a quadratic equation (and honestly one of the ugliest I've seen at this forum

)

There exist a formula to solve quadratic equations:

If the equation is in the form

$ax^2 + bx + c = 0$ then the solutions are:

$x_1=\frac{-b + \sqrt{b^2-4ac}}{2a}~\vee~ x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}$

To your question:

1. Bring your equation into the required form:

$0.01 = -0.027c^2 + 8c - 280~\iff~ -0.027c^2 + 8c - 280.01 = 0$

2. Now plug in the values into the formula:

$x_1=\frac{-8 + \sqrt{8^2-4\cdot (-0.027) \cdot 280.01}}{2 \cdot (-0.027)}~\vee~ x_2=\frac{-8 - \sqrt{8^2-4\cdot (-0.027) \cdot 280.01}}{2 \cdot (-0.027)}$

3. Use a calculator. You'll get:

$x_1 \approx 40.55106148...~\vee~ x_2 \approx 255.7452348...$

**Embalmed** · February 5th 2008, 09:08 PM

OMG, that was so many years ago. It is no wonder I forgot. When I read the formula I could hear it in my head from my teacher so long ago. Thanks so much!

**Soroban** · February 6th 2008, 06:28 AM

Hello, Embalmed!

I was trying to help my daughter solve a problem from her algebra homework.

$0.01 \:= \:-0.027c^2 + 8c - 280$

Answer: . $c \,= \,40.5511$

I would immediately multiply by 1000 ... and get: . $10 \:=\:-27c^2 + 8,000c - 280,000$

Then we have the quadratic: . $27c^2 - 8000c + 280,000 \:=\:0$

Quadratic Formula: . $c \;=\;\frac{8000 \pm\sqrt{8000^2 - 4(27)(280,000)}}{2(27)} \;=\;\frac{8000 \pm\sqrt{33,758,920}}{54}$

And there are two solutions: . $\begin{Bmatrix}255.7452348 \\ \text{and} \\ 40.55106148\end{Bmatrix}$

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