How to Factor Quadratic Equations with Two Variables?

[MrNonexistent]

I forgot How to do these...

Homework Statement

y = x^2 - 8x + 10
change into y = (x[plus or minus]A)*(x[plus or minus]B)

I just forgot how to do these... real fast could someone solve w/ showing work?

Thanks. I just don't remember how to do it...

 

[rocomath]

You need to find 2 numbers that give you the difference of -8 and the product of positive 10.

Since you won't be able to find it by simple factorization, you have to use the quadratic formula.

Google is your friend!

 

[MrNonexistent]

Oh ok thanks.

yea, i was just running factors over and over in my head and i wasn't coming up with anything, I'm kind of brain dead, so i wasn't sure if it was just me. Thanks.

 

[rocomath]

1 and 10
2 and 5

:(

 

[MrNonexistent]

You need to find 2 numbers that give you the difference of -8 and the product of positive 10.

Since you won't be able to find it by simple factorization, you have to use the quadratic formula.

Google is your friend!

when i type in on Google quadratic formula, then i plug the equations in, it tells me x=16 and x = 8

? not the answer i need.

 

[MrNonexistent]

also those were wrong, i did it wrong. hold on

 

[rocomath]

Nope ... try it again.

 

[MrNonexistent]

ok i think i got it, is it just (x - [8+$$\sqrt{24}$$/2]) * (x - [8-$$\sqrt{24}$$/2])?

 

[rocomath]

ok i think i got it, is it just (x - [8+$$\sqrt{24}$$/2]) * (x - [8-$$\sqrt{24}$$/2])?

Yes, but don't forget to reduce!

 

[MrNonexistent]

by that, you just mean plugging it into a calculator and finding the answer, correct?

which happens to be 6.5505 & 1.4495 correct?

 

[rocomath]

by that, you just mean plugging it into a calculator and finding the answer, correct?

which happens to be 6.5505 & 1.4495 correct?

No. $$\frac{8 \pm \sqrt{4 \cdot 6}}{2}$$ can reduce.

 

[MrNonexistent]

so 4 $$\pm \sqrt{2*3}$$?

 

[rocomath]

Yeppp.