anonymous
  • anonymous
need help !! :))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
with?
anonymous
  • anonymous
1 Attachment
phi
  • phi
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

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anonymous
  • anonymous
what am i supposed to d then? how do i solve it
phi
  • phi
You did not post the question. I assume they want to know which line contains a mistake. If so, that would be line 2.
anonymous
  • anonymous
i have to find the mistake then solve it correctly
phi
  • phi
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 \[ x-3 = \sqrt{17-4x} \]
phi
  • phi
now square both sides. on the left that means write (x-3)^2 on the right , what do you get after squaring ? any idea?
anonymous
  • anonymous
X^2-6X+9?
phi
  • phi
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)?
anonymous
  • anonymous
X^2-6X+9=17-4X
phi
  • phi
now add -17 to both sides and +4x to both sides
anonymous
  • anonymous
X^2-2X-8?
phi
  • phi
you mean X^2-2X-8=0 remember it is an equation
anonymous
  • anonymous
A=1,B=-2,C=-8
phi
  • phi
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
anonymous
  • anonymous
OH YEAH
anonymous
  • anonymous
THATS IT? thanks so much!! :)
phi
  • phi
(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
anonymous
  • anonymous
-2 is false. thanks:)
phi
  • phi
yes

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