Can you guys find x?

>>25953544
the square root of 10 is 5, so the other side has to be 5. 5-2 is 3, so X=3. solved

>>25953566
>the square root of 10 is 5
No it isn't.

>>25953597
he's fucking with you anon

>>25953597
5*2=10

checkmait, athiests

>>25953544
>Can you guys find x?
It's under the triangle OP

>>25953544
x^2 + (x+2)^2 = sqrt(10)^2
i'm too lazy to actually find it though, someone else do it

>>25953544

its 1 anon. the 2^2 = 4, with the remainder being 6

so ((X * 2 * 2) + x^x) + x = 6

>>25953642
you guys are funny

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>>25953544

done

veryoriginal.ini

x^2+2x-3=0
(x+3)(x-1)
x=1, since you can discard the negative

>>25953659

4x + 2x = 6
6x = 6
x = 1

(Sqrt (10))^2 =x^2 + (x+2)^2
10 = x^2 + (x+2)(x+2)
10 = x^2 + x^2 +4x + 4
10 = 2x^2 + 4x + 4
5 = x^2 + 2x + 2
0 = x^2 + 2x - 3
0 = (x + 3)(x - 1)
x = -3 or x = 1
Well you can't have a negative length so x = 1?

>>25953659

This. A^2 + B^2 = C^2

You can solve for the hypotenuse if you know what A and B are.

1^2 + (1+2)^2 = [square root of "10"]^2
1 + 9 = 10

Then again it's been years since I've been in school so this is just what I remember.

>doing underage anon's homework

>>25953702
>>25953659

x^2 + [(x* 2*2)+ x^2]+4 = 10
x^2 + [(x* 2*2)+ x^2] = 6
x^2 + [(x*4) +x^2] = 6
2x^2 + 4x = 6

>>25953642
The first step to become a smart ass. Good job.

x^2+(x+2)^2=10
2x^2+4x-6=0
x^2+2x-3=0
(x+3)(x-1)
x=1

>>25953544
x=sqrt10/2
(x+2)=sqrt30/2

OP is a faggot though.

>>25953865
>The first step to become a smart ass. Good job.
Good English m8

>>25953544

We first of all consider that st10 is approximately 3.16. Without any further checks, it immediately /seems/ reasonable that a length and two added to it, may comprise the legs of some right triangle together with st10. But I won't check this since the point right now is for me to work mentally, and write this post.

OTTOMH (and without using paper), The Pythagorean Theorem, together with immediate re-arrangement of the constant term, gives

2x^2 + 4x - 6 = 0

From here, it remains to solve for x, using the quadratic formula, which will produce two results. If either result is anything other than a positive real number (say, a negative number, or a non-real complex number), then we will be obliged to reject it, and OP's construction. Presenting scattered steps-in-progress of the QF gives

(-4 +- st(16+48) ) / 4 = (-4 +- 8) / 4 = 1 OR -3.

We reject the negative result per the above, and conclude that x = 1 (or, -3)!. Moreover, this bears inspection: 1 + 9 = 10. Interestingly, note that the negative case is "complementary", a la a unit circle describing trig results as it moves around a quadrant. In this sense, the negative result is not an absurdity, which does yet sometimes happen when dealing with quadratics in "ordinary" contexts.

( I was eventually obliged to use quick pen-and-paper, sadly, as I'd forgotten the "2a" term in the demoninator and it bugged my autism. But the point is that OP's problem has a literal, real solution, as well as a complementary negative solution, and is not a trick question. )

>>25953544
No. I can't find your x.
I don't know y she left either

>>25954237
>robots doing math

this makes me mad. Also your answer is wrong.

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No cause I'm not a fucking nerd.

>>25953544
tell your teacher he's a retard. all the trig you need for calculus are basic identities.

>>25954237
the negative is not complementary, stop misleading him

>>25954324
yu r retarded

We know a^2+b^2 = c^2

So we have:
x^2 + (x+2)^2 = 10
Which, when factored out is:
2x^2 + 4x + 4 = 10
Then move it all to the same side:
2x^2 + 4x - 6

2 (x^2 + 2x - 3)
2(x+3)(x-1)

X = 1
X = -3

Since it's a triangle, the numbers must be real, so X = 1

Is this correct?

>>25953745
Yup, this guy got it

>>25954375
>real
Didn't mean real, meant positive

(x+2)^2 + x^2 = 10
If x is 1 that would be (1+2)^2 + 1^2 = 10
3^2 + 1^2 = 10
9+1=10
x=1

>>25953942

except that's wrong, you fucking retard.

This is literally first year of highschool/maybe junior high
Post something different

>>25953745
This one is correct.

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>>25954262

Of course it isn't, you objectively stupid fucking idiot. the 1 is manifestly correct, as multiple other anons have agreed, as even you should know, you objectively stupid fucking idiot. Therefore, you objectively stupid fucking idiot, what remains is to consider the degenerate negative case, you objectively stupid fucking idiot.

And it can obviously be accounted for in a creative and mathematically legitimate fashion. Although of course lengths, a la absolute values, are manifestly non-negative, no one bristles when the "negative" coordinates of the unit circle are introduced in babby trig. And so it is with this other item. However one wishes to render it (there are multiple possibilities), it is possible to treat of the other quadratic case something like this.

>>25954322

Certainly it is, you just have to think a bit, as I've shown. Your "him" is unclear.

>>25954637
The lower hemisphere hypotenuse should be: -sqrt(10).

https://www.youtube.com/watch?v=PdYunqCBRbo

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>>25953544
here you go goi, everyone above this post is beyond retarded

off yourself retard vegetables

X = -3 and X = 1

fuck off retards, proof in the pic

a^2 + b^2 = c^2
x^2 +(x+2)^2 = 10
x^2 + x^2 +4x +4 = 10
2x^2 +4x - 6 = 0
x^2 +2x - 3 = 0
x = -3, 1

Throw away -3 since impossible

Answer is 1

>>25954786
Nice imaginary number fuckboy.

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Everyone can factor polynomials, fuckbois.
Go back to posting smug anime grills

>>25955065

You are also a stupid person, and it pains me to think that I might be confused with you.

-st(10) = -3.16...

st(-10), on the other hand, is what you had in mind as your imaginary number, being i * 3.16.... Two different things.

>>25955000

Like the others, you are right in that you identify the obvious and correct answer, via correct method. However, -3 is also an "extended" answer, and not at all impossible, nor any common-core fuckery.

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>>25953566
Ghjjhtyjjtth