Question

need help !! :))

Answer 1

with?

Answer 2

line 1 is ok.
but line 2 is wrong
in other words,
(x-y)^2 is *not* x^2-y^2 (which is what they did)
(x-y)^2 is x^2 +y^2 -2xy

Answer 3

what am i supposed to d then? how do i solve it

Answer 4

You did not post the question. I assume they want to know which line contains a mistake. If so, that would be line 2.

Answer 5

i have to find the mistake then solve it correctly

Answer 6

ok.
line 1 is technically not a mistake, but it is the wrong first step
(line 2 is definitely a mistake)

I would start over, with the original
\[ x - \sqrt{17-4x}= 3 \]
and put the square root on the right-hand side , like this:
\[ x - \sqrt{17-4x} +\sqrt{17-4x} = 3 + \sqrt{17-4x} \]
the left side simplifies
\[ x= 3 + \sqrt{17-4x} \]
now add -3 to both sides
\[ x-3 = \sqrt{17-4x} \]

Answer 7

now square both sides. on the left that means write (x-3)^2
on the right , what do you get after squaring ? any idea?

Answer 8

X^2-6X+9?

Answer 9

yes, that is the left side. the right side is "easy" --- squaring gets rid of the square root side.
can you write down the whole equation (both sides)?

Answer 10

X^2-6X+9=17-4X

Answer 11

now add -17 to both sides and +4x to both sides

Answer 12

X^2-2X-8?

Answer 13

you mean
X^2-2X-8=0
remember it is an equation

Answer 14

A=1,B=-2,C=-8

Answer 15

yes. you can use the quadratic formula
(if they want you to)
but you can factor this: notice -4* 2= -8 and -4+2= -2

Answer 16

OH YEAH

Answer 17

THATS IT? thanks so much!! :)

Answer 18

(x-4)(x+2)= 0
either x-4=0 --> x=4
or x+2 =0 ---> x= -2

however, with these types of problems you have to check both solutions.
sometimes you get an "extraneous" solution. So check both values in the original equation

Answer 19

-2 is false. thanks:)

Answer 20

yes