Question

SOLVE THE EQUATION BY COMPLETING THE SQUARE. X SQUARE +8x-33=0

Answer 1

\[x ^{2}+8x-33\]

Answer 2

First, take the number infront of x, divide it by 2, and then square it.

Answer 3

16

Answer 4

Alright, now add and then subtract it after 8x so you have:
\[x ^{2}+8x+16-16-33\]

Answer 5

and then factor the \[x ^{2}+8x+16\]

Answer 6

Hint: itll be (x-a)^2 where a equals the number you divided by 2 before you squared it

Answer 7

so 4

Answer 8

ok

Answer 9

So now we have (x+4)^2-16-33

Answer 10

Move the numbers to the right and you get
(x+4)^2=49

Answer 11

What would you do next to solve for x?

Answer 12

ok -49 add

Answer 13

wait -4

Answer 14

What?

Answer 15

im lost

Answer 16

If you're solving for x you would take the square root of both sides. it cancels the square out on the left so we have

\[x+4=\pm \sqrt{49}\]

Answer 17

then it would be7, -1?

Answer 18

square root 7

Answer 19

No, now you subtract the 4 from both sides and end up with:

\[x=-4\pm7\]

Answer 20

Since 7 is the square root of 49.

Answer 21

ok

Answer 22

Since its plus or minus, you have two separate answers.

x=-4+7 and x=-4-7
or
x=3 and x=-11

Answer 23

But when you take a square root, you have to go back and check your answers. So plug 3 in for x in the original equation and if its true, that is an answer. Same for -11

Answer 24

ok that would be an easier way of checking.

Answer 25

i have another question

Answer 26

Did you check the answers? in this case both 3 and -11 are solutions.


Post your question as a new one and i will look at it

Answer 27

ok