Question

10 points!! Please help! I have been stuck on this for about 7 hours now.





Complete the square on the following quadratic.

1. x^2 + 16x + 9 = 0 and 2. x^2 + 16x + 7 = 0

what value is added to both sides of the equation?

Answer 1

x² + 16x + 9 = 0
x² + 16x + 9 + 55 = 0 + 55   add 55 to both sides
x² + 16x + 64 = 55
(x+8)² = 55


x² + 16x + 7 = 0
x² + 16x + 7 + 57 = 0 + 57   add 57 to both sides
x² + 16x + 64 = 57
(x+8)² = 57


Explanation:
(x+b)² = x² + 2xb + b²
For the first equation:
2xb = 16x and you need to find b²
If 2xb=16x then b=16x/2x = 8  and b² = 8² = 64
64-9 = 55
So, you must add 55 to both sides of the equation to complete the square.

Answer 2

1) x²+16x+9 =0 → (x²+16x) +9 =0
Complete the square of the parenthesis: x²+16x.
The missing part is a square: x²+16x +?²
We know that 16x is twice the square root of the 1st (that is x) times twice the 1st (that is x) by the 2nd (that is "?") :
16x = (2x).(?) and "?" = 8.
If "?" = 8 its square "?²" = 64
Hence: x²+16x + 64 - 64 +9 = (x²+16x+64) -64 + 9 = (x+8)² - 55
Same logic for the second and you will find:
x² + 16x + 64 - 64 + 7 = (x+8)² - 57
For both equation we have added 64 ( and after subtract it to keep the equilibrium of both equations)