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/sci/ - Science & Math

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2933143 No.2933143 [Reply] [Original]

will a generous anon help a collegefag out? I can't seem to solve this problem. I passed algebra in high school but I cant seem to remember how to do this. Can an anon show me how?

>> No.2933157

1

>> No.2933159

wolframalpha.com

>> No.2933161

square both sides.
Collect like terms
Square again to get rid of the last square root.
It will be messy but that should get it done

>> No.2933167
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2933167

>>2933161
>>2933161
>Square again to get rid of the last square root.

what?
no.

>> No.2933164

<span class="math">3x + 1 = (4 - \sqrt{2x+2})^2[/spoiler]
<span class="math">3x + 1 = (4 - \sqrt{2x+2})(4 - \sqrt{2x+2})[/spoiler]

Multiply out and solve for x, nigga.

>> No.2933175

>>2933164

I'll give you some more help.

<span class="math">3x + 1 = 18 + 2 x - 8\sqrt{2 + 2x}[/spoiler]

>> No.2933176

i remember these. needlessly tedious. square both sides. this gets rid of one of the radicals. move the remaining radical to one side, the other terms to the other. square again to remove second radical. solve for x. its casual shit, just takes longer than it should

>> No.2933178

If x is real, then the RHS is not negative.

This implies sqrt(2x + 2) < 4 => x < 7

Since x > 0 and x < 7, x can be 1, 2, 3, 4, 5, or 6.

Checking all of these only shows x = 1 as the answer.

>> No.2933184

>>2933167
What happens when you square (4- sqrt(2x + 2))?
You are still left with a term with a square root in it.

>> No.2933185

>>2933176
Exactly why I was reluctant to work it all out. 1) i'm lazy 2) I dont expect this shit in an intermediate algebra class...fuck I want to be back in calc. thanks anon

>> No.2933199

>>2933185
>Wants to be in calc
>Can't solve this shit

Ouch.

>> No.2933202

>>2933185
All through highschool I am told calculus will be hard as shit and that all the rest of my maths will be easy compared to it. I get to uni and calculus is a joke while algebra and number theory are fucking ridiculous amounts of work.

>> No.2933211

>>2933175
after this,
17-x=8sqrt(2(x+1))
17^2-34x+x^2=128x+128
then just quadratic formula it.

>> No.2933219

what kind of crappy college you are going where they give you so easy problems?

>> No.2933317

Heres a slightly more interesting question.

4 - sqrt(2x + 2) = sqrt(3x + 1)
therefore
x^2 - 162x + 161 = 0
therefore
x = 1 or 161
but x = 1 is the only answer that satisfies the original equation

Where did the x = 161 solution come from, /sci/?

>> No.2933357

>>2933317
"x = 1 -> x^2 = 1 -> x = 1 or -1"
same way -1 shows up as a "solution" here

>> No.2933359

>>2933317
x^2 - 162x + 161 = 0
(x-161)(x-1)=0
Therefore
X-161=0 ,X-1=0
Therefore
X=161, X=1

>> No.2933376
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2933376

Are you asking why a quadratic has two solutions?

More importantly, are you asking why (x - 1)(x - 161) equals the original equation?

>> No.2933381

16 - 8(2x+2)^(1/2) + (2x+2) = 3x + 1
8(2x+2)^(1/2) = 16 + 2x + 2 - 3x - 1
64 ( 2x + 2 ) = (17-x)^2
128x + 128 = 289 - 34x + x^2
0 = x^2 - 162x + 161
0 = (x-161)(x-1)
x = 161
x = 1

>> No.2933394

I decided to start plugging in numbers starting with 0 and was pleasantly surprised.