Algebra help?

Question

p^2 +(p-8)^2= 130

Please help me with this problem. A step-by step explaination would be awesome so I understand how to do it next time. Thanks!

^2 is squared. I wasn't sure how else to write it.

Answer 1

=>p^2+p^2-2*8*p+8^2=130
2p^2-16p+64=130
take common factor 2 and cancel it..
p^2-8p+32=65
p^2-8p-33=0
p^2-11p+3p-33=0
p(p-11)+3(p-11)=0
(p+3)(p-11)=0
p= -3 or p= 11..
hope dat helped
jj

Answer 2

I would multiply out p-8 * p-8
then collect like terms p^2

p^2 + (p^2 -16p + 64) = 130

2p^2 -16p -66 = 0 [collect both p^2-terms and constants]

p^2 - 8p - 33 = 0 [reduce by dividing coefficients by GCF 2]

(p + 3)(p - 11) = 0

p = -3 or p = 11
check your work by plugging into
ORIGINAL printed question
see if both = 130

(-3)^2 + (-3-8)^2 = 9 + 121 = 130 this works
(11)^2 + (11-8)^2 = 121 + 9 = 130 this works

Answer 3

It'll probablly be wrong, but I'll give it a shot:

p^2 + (p-8)^2 = 130

p^2 + (p+8)(p-8) = 130

p^2 + p^2-8p+8p-8 = 130

p^2 + p^2 - 8 = 130

2p^2 = 130

p^2 = 69

p = square root of 69

Answer 4

so, first you would square everything in the parenthesis

p^2 + p^2-16=130

then, combine like terms

2p^2-16=130

divide everything by 2 to isolate p

p^2 - 8 =65

subtract 8 from both sides

p^2=57

take the square root of both sides

p=7.548.....

either this answer or the p^2=57 depending on how your teacher wants you to write it.

good luck!

hmmmmm.......her answer looks better above me!