(x+9)^2-32=0 solve and simplify?

Question

(x+9)^2-32=0
(x^2+18x+81)-32=0
x^2+18x+81-32=0
x^2+18x+49=0

Where x is a real number
I do not know which part of my
solution is where I find what x=

If there is more than one solution separate with commas

Answer 1

This particular quadratic equation will require the quadratic formula:

-b +/- sqrt(b^2-4ac) all divided by 2a.

Here, a=1, b=18, and c=49

-18 +/- sqrt[(18^2 - 4(1)(49)] all divided by 2.

18^2 - 4 x 49 = 128.....

Let's do the top first:

-18 +/- sqrt(128)

What is sqrt(128)? You need to factor 128 = 64 x 2
Therefore, sqrt(128) = sqrt(64 x 2) = 8 x sqrt(2).
(we took 64 out of the root function, so comes out as the sqrt, which is 8.)

Set it all up:

-18 +/- 8sqrt(2) divided by 2:

-9 +/- 4sqrt(2)

Answer: -9 - 4sqrt(2), -9 + 4sqrt(2).

Not easy due to the answer not a set of integers.

Answer 2

is it x+5 or x-5 regardless of if this is x+5 then: ( x + 5 ) = ( ?x + 7 ) sq. the two facets: x^2+10x+25 = x+7 make it equivalent to 0: x^2+9x+18=0 (x+6)(x+3)=0 x=-6 or x=-3 regardless of if this is x-5 then: ( x - 5 ) = ( ?x + 7 ) x^2 - 10x + 25 = x+7 x^2 - 11x + 18 = 0 (x-9)(x-2)=0 x=9 or x=2

Answer 3

(x+9)^2-32=0
is really equal to
x^2+18x+49=0
D = "b^2 - 4ac" = 18^2 - 4*1*49 =128 = 2 * 64
x(-) = (-18 - 8Sqrt(2))/2 = -9 - 4Sqrt(2)
x(+) = (-18 + 8Sqrt(2))/2 = -9 + 4Sqrt(2)

Answer 4

(x+9)^2-32=0
(x+9)^2 = 32
x+9 = + -sqrt(32)
then
x= -9 + -sqrt(32)
s

Answer 5

x^2+18x+49=0
x=(-18+-Sqrt[324-(4*1*49)])/2
x=-9+-Sqrt[324-196]/2
x=-9+-Sqrt[128]/2
x=-9+-Sqrt[2^7]/2
x=-9+-8Sqrt[2]/2
x=-9+-4Sqrt[2]
x=-9+4Sqrt[2], -9-4Sqrt[2]

Answer 6

O_o