2 problems I do not understand.. Please help explain..?

Question

1. Solve:
x^2 + 6 = 31

2. Solve:
-9(x - 3)^2 = -7

Answer 1

x^2 + 6 = 31
x^2 =31-6 = 25
x = +/- sqrt(25) = +/- 5

2. Solve:
-9(x - 3)^2 = -7
(x-3)^2 =7/9
x- 3 =+/- sqrt(7/9)
x = 3 +/- (1/3)sqrt(7)

Answer 2

All u gotta do is solve for "x" for both the problems so here we go:

1) x^2 + 6 = 31

x^2 + 6 - 6 = 31 - 6

x^2 = 25

Take the square root of both sides, therefore u get:

x = +/- 5 (The reason we write +/- is because if u multiply -5 with -5, u still get 25 as an answer. Therefore "x" equals both plus or minus 5)

2) First divide -9 on each side:

-9(x-3)^2 = -7

(x-3)^2 = -7/-9

(x-3)^2 = 7/9 (since minus and minus cancel and makes plus 7/9)

Now take the square root of both sides and solve for x:

sqr.root (x-3) ^ 2 = sqr. root 7/9

And u get:

x - 3 = (sqr.root 7)/3 (because sqr.root of 7 is a decimal and sqr. root of 9 is simply 3, so u just write it like that!)

Now add 3 on both sides and u get:

x = +/- sqr.root 7/3 + 3 (Same deal here)

And thas ur answer right there!

Answer 3

X^2= 31-6=25
x=5

Let (x-3) = y
y^2=7/9 & y= +or- sq rt 7/9
& x-3 =same
& x=3 +or - sq rt 7/9

Answer 4

1)
x^2 + 6 = 31
x^2 =31-6
x= 25
x = +/- sqrt(25)
= +/- 5

2)
-9(x - 3)^2 = -7
(x-3)^2 =-7/-9
(x-3)^2 =7/9 ---- minus and minus=+
x- 3 =+/- sqrt(7/9)
x = 3 +/- (1/3)sqrt(7)
x=+/-((sqrt7)-9)/3

Answer 5

x^2 = (31-6) = 25
x=+/- 5

(x - 3)^2 = -7/-9
x-3 = sqrt(7/9)
x= 3 +/- sqrt (7/9)

Answer 6

31-6=25

Square root of 25 is +5 and -5

x equals the (plus or minus square root of 7 over 3) plus 3

Answer 7

x^2=31-6
x=squr root of 31-6