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I'm hoping for link to some resource which can explain why the following is true.

$$ x^2 + 104x - 896 = 0 $$

Using the quadratic formula we pull a = 1, b = 104, c = 896. Putting that into the formula for the discriminant we get $ 104^2 - 4.1.896 $ which is 10816 - 3584 = 7232.

Using the quadratic formula the discriminant is 7232 and using the quadratic formula the answer is

$$ -104 \pm \sqrt{7232} \over 2 $$

This simplfies to $ -72 \pm 4 \sqrt{113} $.

The problem I have is I have not found anything which explains why if you plug x = 8 into the equation it also balances out. What I have found out infers x = 8 being an answer to the equation but not a factor and so not a solution but I don't really get the point.

Any links that explain the distinction would help. I have an infinite number of these equations which I'm looking for integer answers to so if said link also pointed out how you can obtain the integer answers like the 8 in this case instead of the irrational provided by the quadratic formula that would be great.

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    $\begingroup$ Scrap that 896 is correct. 8^2 is 64. 8*104 = 832 so it is 896. Checking the quadratic to see if I made a mistake working that out instead. I have other examples anyway. $\endgroup$
    – Rhuaidhri Tynan
    Commented Dec 9, 2013 at 11:12
  • $\begingroup$ @Jp McCarthy sorry editied comment instead of answering again. I've double checked and I had an error alright but it was in the discriminant. Have changed to correct value. 8 still works but the value provide by quadratic formula had the typo. I'll write out all values so you can double check and point out the error if existant. $\endgroup$
    – Rhuaidhri Tynan
    Commented Dec 9, 2013 at 11:27
  • $\begingroup$ @Jp McCarthy have written out all the calculations please spot the error for me please. If it is only down to a bad calculation on my part rather than an lack of understanding on roots versus answers which work when plugged in I'll be happy. $\endgroup$
    – Rhuaidhri Tynan
    Commented Dec 9, 2013 at 11:42

3 Answers 3

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The quadratic formula is $$x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}.$$ Thus in your problem, $$x = \frac{-104\pm\sqrt{104^2 - 4(-896)}}{2} =\frac{-104 \pm\sqrt{14400}}{2} = \frac{-104 \pm 120}{2}.$$ This gives $x = 8$ or $-112$.

It seems the problem in your computation was you forgot the negative on $896$ when you plugged it into the quadratic formula.

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  • $\begingroup$ lols thanks. I must have done that multiple times. Some of the stuff I read earlier tonight wasn't very clear was (it was saying something about plugging values in for specific terms which made me think there was a lack of understanding on my part instead of just bad sums. $\endgroup$
    – Rhuaidhri Tynan
    Commented Dec 9, 2013 at 12:02
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I cannot comment, but you are obviously mistaken in your calculation. Your quadratic formula that is. Here, WA does not make mistakes

EDIT: $$\frac{-104\pm\sqrt{104^2+4\cdot 896}}{2}\\ =\frac{-104\pm\sqrt{14400}}{2}=-52\pm60$$

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  • $\begingroup$ ndh I have read quite a bit trying to find the answer. Please check my values which I have written out now in full and point to the mistake. I've tripled checked this example and I have plenty of others. As I said it isn't that the quadratic formula doesn't give the correct value, all the values work out if you use them for x. $\endgroup$
    – Rhuaidhri Tynan
    Commented Dec 9, 2013 at 11:38
  • $\begingroup$ I added the correct use of the quadratic formula. There are only two answers here $\endgroup$
    – Flowers
    Commented Dec 9, 2013 at 11:50
  • $\begingroup$ He made two mistakes: (1) is that '-104/2' is '-52' not '-72' (2) is that "1042−4.1.896" ("." is '*') should be '1042+4*896' $\endgroup$
    – Flowers
    Commented Dec 9, 2013 at 12:05
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obviously 8 is one of the solutions with no doubts at all.

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