quaudratic functions: solving x^ - 6x - 16 = 0

[Missy]

I haven't done algebra for over thirty years, and I can not get the just of it. I hope that you can help me.
The problem is x^-6x-16=0. I need to know it I am on the right track with this problem

x^-6x-16=0
x^-2x-4=0 Can you tell me if I am on the right tack or if I am completely doing this wrong.
(x-2)=0 (x-4)=0
x-2+2=0+2 x-4+4=0+4
x+0+2 x-4+4=0+4
x=2x x+0+4
x=4

 

[Loren]

Re: quaudratic functions

Is there a typo?
x^-6x-16=0 is not a meaningful equation. Or do you mean x^2-6x-16=0?

If that is what you mean...

\(\displaystyle x^2-6x-16=0\)
(x - 8)(x + 2) = 0
x-8=0 or x+2=0
If x-8=0 then x=8.
If x+2=0 then x=-2.
Check:
If x=8 ---> \(\displaystyle x^2-6x-16=0 ---> 8^2-6\cdot 8-16=0 ---> 64 - 48 - 16 = 0\). It checks.
If x=-2 --->\(\displaystyle (-2)^2-6(-2)-16=0 ---> 4 + 12 - 16 = 0\). That, also, checks.

 

[stapel]

x^ - 6x - 16 = 0
x^ - 2x - 4 = 0

Note: I will guess that "x^" is meant to be "x^2" (that is, "x squared" or "x[sup:3pqsm650]2[/sup:3pqsm650]").

What was your reasoning behind subtracting 4x and 14 from only one side of the equation? (Since this gives you an entirely new equation with new solutions, I'm not sure what you're accomplishing with this...?)

Please be complete. Thank you!

Eliz.