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. )
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.
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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