What are the values of x in the equation x2 – 6x + 9 = 25? x = –2 or x = 8 x = –1 or x = –11 x = 1 or x = 11 x = 2 or x = –8

The answer is A, x= -2 or x= 8

Step-by-step explanation:

x = –2 or x = 8

Step-by-step explanation:

x²-6x+9 = 25

Subtracting 25 from both the sides,

x²-6x+9-25 = 25 -25

Cancelling out 25 and -25 from the right side

x²-6x+9-25 = 0

1x²-6x-16 = 0

Multiplying the quotient of the first term with the last term,

1*(-16) = -16

We need to find out two such factors of -16 which when added gives us the middle term coefficient -6.

-8 and 2 are such factors of -16, which when added gives us the middle term coefficient -6.

So, we rewrite the middle term -6x as -8x+2x

x²-8x+2x-16 = 0

Factor out x from the first two terms

x(x-8) + 2x-16 = 0

Factor out 2 from the last two terms

x(x-8) + 2(x-8) = 0

Factor out (x-8) from both the terms

(x-8)(x+2) = 0

Either x-8 = 0 or x+2=0

Solving x-8=0, we get x = 8

Solving x+2=0, we get x= -2

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

it shoild be o

step-by-step explanation: