How the HECK to I solve this algebra equation?

Question

Solve the equation:
p^2 + (p + 9)^2 = 45

Any idea as to how to do this? Please help!

Thank you so much!!!

Answer 1

Use the formula (a+b)^2 = a^2 +b^2+2ab

So,
p^2+(p^2+9^2+2*p*9) = 45
p^2+p^2+81+18*p = 45
This gives: 2*p^2+18*p+81-45=0
or: 2*p^2+18*p+36=0
Taking 2 common from the left hand side yields:

p^2+9*p+18=0
p^2+3*p+6*p+18=0
p(p+3)+6(p+3)=0
p+3=0 , p+6=0

The equation has 2 roots;
p= -3 , p= -6

Answer 2

you need to solve th brackets first.

p^2 +p^2 + 81+18 p = 45 solving the bracket

2p^2 + 18p = 45 - 81 subtracting 81 from both sides

2p^2 + 18p = -36

2p^2 + 18p + 36 = 0 adding 36 on both sides

now solve for the quadratic equation.
(p + 3)(p + 6) = 0

p = -3
p = -6

Answer 3

p^2 + p^2 + 18p + 81 = 45

2p^2 + 18p + 36 = 0

p^2 + 9p + 18 = 0

(p + 6) (p + 3) = 0

p = -6, p = -3

Answer 4

i used my calculator to solve this, here is what i got:

p^2 + (p + 9)^2 = 45
i put in:
Solve(p^2 + (p + 9)^2 = 45,p)
answer: p= -6 or p= -3 ( p= negative six or negative three)

hope its correct :) lol

Answer 5

Figure out what (p+9)^2 is first.
combine like terms.
You'll end up with a quadratic equation.
Solve the quadratic equation