• Sep 3rd 2009, 08:11 PM
I'm working on basic problems to review for a placement test into calculus. The questions are here, I don't need you to click on if your not interest, they are multiple choice. College Algebra

Where am I going wrong here?
$\displaystyle x^2-2x-24=24$
$\displaystyle x^2-2x-48=0$
$\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(-48)(1)}}{2}=\frac{2\pm\sqrt{4+182}}{2}=\frac{2\pm\ sqrt{186}}{2}$

The solution sets (in the multiple coice) are all whole numbers. I have$\displaystyle \pm\frac{\sqrt{186}}{2}\approx6.819$

Where am I going wrong here?
• Sep 3rd 2009, 08:25 PM
Rapha

Quote:

I'm working on basic problems to review for a placement test into calculus. The questions are here, I don't need you to click on if your not interest, they are multiple choice. College Algebra

Where am I going wrong here?
$\displaystyle x^2-2x-24=24$
$\displaystyle x^2-2x-48=0$
$\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(-48)(1)}}{2}=\frac{2\pm\sqrt{4+182}}{2}=\frac{2\pm\ sqrt{186}}{2}$

The solution sets (in the multiple coice) are all whole numbers. I have$\displaystyle \pm\frac{\sqrt{186}}{2}\approx6.819$

Where am I going wrong here?

This is because 4*48 = 192 not 182

Yours
Rapha
• Sep 4th 2009, 04:25 AM
HallsofIvy
You might also have noted that 6(8)= 48 and 8- 6= 2. That is, $\displaystyle x^2- 2x- 48$ factors as $\displaystyle (x- 8)(x+ 6)$.