solve x^2 = 8x -16: which method would you suggest?

Re: solve x^2 = 8x -16

Put your equation into ax[sup:1ltl5k2j]2[/sup:1ltl5k2j] + bx + c = 0 form, then factor the left side and go from there.
 
Re: solve x^2 = 8x -16

x^2 = 8x - 16
ginny1029 said:
which method would you suggest?
Click to expand...
Ginny, what grade are you in? Or are you doing maths on your own?
 
Re: solve x^2 = 8x -16

wow Denis am i that bad lol
i am in elementary algebra in college
 
Re: solve x^2 = 8x -16

Loren I tried this and got lost and then assumed I wasnt doing something right but here goes...

x^2 = 8x + 16

x^2 -8x - 16 = 0

this looks to be a perfect square

ie

(x + 4) ^2 but this is were i come to a road block
 
I responded to what you indicated in your original post. The equation was x[sup:dlwpxiwb]2[/sup:dlwpxiwb]=8x-16.
If you translated this into x[sup:dlwpxiwb]2[/sup:dlwpxiwb]-8x-16=0, you made a mistake.
 
im sorry i dont know why my problem is but if i start over

i could do the x^2 -8x = -16
then find what number would fit here

x^2 - 8x +_____ = -16

x^2 -8x + 16 = -16

(x - 4) squared = -16

x - 4 =the sq rt of -16 you cant have a sq rt of a neg number so im back to being lost
 
Don't you have to add that squared term to both sides of the equation...? :wink:
 
So on the left side you have a binomial squared and on the right side you have zero. That's the whole idea. You want to have an equation where you multiply two (or more) things together and get zero. If that is the case, then you know that one of those "things" has to be zero. The only way you get zero is by having one of the factors equal to zero. For instance, 3 times 0 equals 0. 3 times anything else cannot be zero. So, if you have something like (x-4)(x+2)=0 you know that either x-4 is zero or x+2 is zero. If x-4=0 then you know that x=4. If x+2=0, then you know that x=-2.
 
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