This isn't really much of an algebra problem but different people are giving me different interpretations of " Square of One more than the Larger Part ". I've been given

" Separate 17 into two parts. So that the sum of their squares is equal to the "SQUARE OF ONE MORE THAN THE LARGER PART". Find the Smaller Part. "

I interpreted it myself; and I got

$x^2 + y^2 = (y^2 + 1)^2 = 17$

I don't think this works. I feel I have misinterpreted the problem. Other people have other interpretations too but they don't work either.

The correct answer is 5.

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    $\begingroup$ I don't read it the same way at all: The square of ...l so $(...)^2$. ...Of one more than the larger part: $y + 1$. So it should be $(1 + y)^2$. If it was the way you wrote it, it would be "square root of one more than the square of the larger part." $\endgroup$ – Jared Jun 3 '15 at 0:52
  • $\begingroup$ whoops sorry. yep what i meant should be like that i copied wrongly but the question still stands! $\endgroup$ – james Jun 3 '15 at 0:53
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    $\begingroup$ Your two equations should be $x+y=17$ and $x^2+y^2=(y+1)^2$ You could subtract out a $y^2$ from the second equation after expanding. $\endgroup$ – turkeyhundt Jun 3 '15 at 0:55
  • $\begingroup$ Wait lol you totally right. it should be (y+1)^2 ! $\endgroup$ – james Jun 3 '15 at 0:55
  • $\begingroup$ You have $x+y = 17 \rightarrow x = 17 - y$, plug into the original equation (which your's is still wrong), and solve for $x$. You should get two solutions--which represent the values of $x$ and $y$. $\endgroup$ – Jared Jun 3 '15 at 0:55

"Separate $17$ into two parts."

That tells me you should set up $x+y = 17$ (equation 1)

"The sum of the squares of the parts is equal to the square of one more than the greater part."

We can let $y$ be the greater part without any loss of generality (which makes $x$ the smaller part). That means that $x^2 + y^2 = (y+1)^2$. (equation 2).

Now you have to solve the two equations simultaneously.

Start by manipulating equation 2:

$x^2 = (y+1)^2 - y^2 = 2y+1$

$y = \frac{1}{2}(x^2 - 1)$

Substitute that expression for $y$ into equation 1 to get:

$x + \frac{1}{2}(x^2 - 1) = 17$

After rearrangement, you should get the quadratic:

$x^2 + 2x - 35 = 0$

which can be factorised:

$(x+7)(x-5) = 0$

So $x=-7$ or $5$.

Here we generally discard the negative root as splitting a number into two parts generally refers to natural number parts. Perhaps this needs better clarification in the question. But assuming this is the case, the only admissible solution is $x=5$.

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