Welcome to our community

Be a part of something great, join today!

May be a Quadratic Problem

Casio

Member
Feb 11, 2012
86
Hi everyone(Wink)

I have an equation which looks like quadratic to me.

2(x - 5)2 = 32

Can I divide both sides by 2?

(x - 5)2 = 16

Can I take the square root?

x - 5 = + or - 4

if x = 5 plus 4 = 9

if x = 5 - 4 = 1

Therefore my roots would be;

x = 9 or x = 1

Is this method acceptable and the correct way to find the values of x. I know it works out OK but am wondering about the way I have done it?

Thanks
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Hi Casio!

(Yes) Well done. That's exactly how you should do this problem every step of the way.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Yes, that is a perfectly valid procedure. All of your steps are correct. (Nod)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
After dividing by 2, we may also arrange as the difference of squares:

(x - 5)2 - 42 = 0

((x - 5) + 4)((x - 5) - 4) = 0

(x - 1)(x - 9) = 0

x = 1, 9