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I am trying to solve this equation using the quadratic formula:

$$x^2 + 4x -1 = 0$$

I start by substituting the values into the quadratic formula:

$$x = {-(4) \pm \sqrt {(4)^2 - 4(1)(-1)} \over 2}$$

which becomes

$$x = {-4 \pm \sqrt{20} \over 2}$$

This is the answer the textbook that I am using gives but I would have thought I could have simplified this further to:

$$x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$$

which becomes

$$x = {-4 \pm 2 \sqrt 5 \over 2}$$

which becomes

$$x = -2 \pm \sqrt 5$$

Am I right and if so, why would the textbook not have simplified it further?

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    $\begingroup$ $$-2\pm \sqrt { 5 } $$ not -4 $\endgroup$ – haqnatural Aug 10 '15 at 20:07
  • $\begingroup$ I'm surprised at how many of the answers are "here, let me solve and simplify that for you'. $\endgroup$ – Hurkyl Aug 11 '15 at 11:13
  • $\begingroup$ @Hurkyl haha, you actually answered my question. I guess these forums are a chance for people to showcase their skills. $\endgroup$ – dagda1 Aug 11 '15 at 13:02
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People often have trouble checking their answers if they aren't in exactly the same form.

By presenting the answer in the form

$$ x = {-4 \pm \sqrt{20} \over 2} $$

students who have at least applied the quadratic formula correctly will not have trouble checking the result. Also, seeing the result in this form might help someone who is having trouble recognize what to do.

Students who go on to simplify this are probably able to check their answer either way.

Of course, I'm speculating as to the actual motivation of the textbook authors. And I do not intend to agree or disagree with the premise that the answer should have been given as $x = -2 \pm \sqrt{5}$.

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Note that

$$\frac{B+C}{A}=\frac{B}{A}+\frac{C}{A}$$


$$x=\frac{-4\pm 2\sqrt{5}}{2}=\frac{-4}{2}\pm\frac{2\sqrt{5}}{2}=-2\pm\sqrt{5}.$$

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  • $\begingroup$ where did ${B + C \over A}$ = ${{B \over A} + {C \over A}}$ come from, I've not seen this before $\endgroup$ – dagda1 Aug 11 '15 at 7:30
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    $\begingroup$ @dagda1 It's the distibutive property of division over addidion. From arithemetic. $\endgroup$ – Taemyr Aug 11 '15 at 9:04
  • $\begingroup$ This answer was written because OP first wrote $x=-4\pm\sqrt{5}$ as the answer. (OP edited it later) $\endgroup$ – mathlove Aug 11 '15 at 11:20
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The classical formula for the quadratic equation is

$$ax^2+bx+c=0\iff x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

But there are many circumstances where the quadratic coefficient is $1$ and the first degree coefficient has an explicit factor $2$, and it is worth to remember the simplified formula $$x^2+2bx+c=0\iff x=-b\pm\sqrt{b^2-c}.$$


In your case, $$x=-2\pm\sqrt{2^2-(-1)}.$$

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Note in this case using the quadratic formula is silly, as it is so easy to complete the square (where the quadatic formula comes from): $$ x^2+4x-1=(x+2)^2-4-1=0\iff (x+2)^2=5\iff x=-2\pm\sqrt 5.$$

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  • $\begingroup$ You got there before I did with exactly the same idea. $\endgroup$ – Mark Bennet Aug 10 '15 at 20:14
  • $\begingroup$ Did I? It was a matter of a couple of minutes. It happens at times. Nice probability problem… $\endgroup$ – Bernard Aug 10 '15 at 20:16
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    $\begingroup$ the question in the textbook wanted the solution to use the quadratic formula not the best way. $\endgroup$ – dagda1 Aug 10 '15 at 20:51
  • $\begingroup$ In this case, as the coefficient of $x$ is even, at least use the reduced formulae: if the equation is $ax^2+2b'x+c=0$, set $\;Delta'=b'^2-ac$. If $\Delta'>0$, the roots are $\; \dfrac{-b'\pm\sqrt{\Delta'}}a$. $\endgroup$ – Bernard Aug 10 '15 at 20:55
  • $\begingroup$ @dagda1 Always good to have a simple way of checking an answer, whatever method you are supposed to use. $\endgroup$ – Mark Bennet Aug 10 '15 at 20:55
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Correct me if I'm wrong but I don't think you are wondering how the authors got from

$x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$

to

$x = -2 \pm \sqrt 5$

(if so, mathlove already answered that).

But rather (or at least I personally would wonder) why the authors start with $x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$ in the first place.

Obviously they want to reduce it. However what you did is a lot more straight forward. Perhaps if they did it this way it would be easier to understand:

$x = {-4 \pm \sqrt{20} \over 2}$

first distribute:

$x = {-4 \over 2} \pm {\sqrt{20} \over 2}$

$x = -2 \pm {\sqrt{20} \over 2}$

now factor what's under the square root, if possible, into one or more factors that are a complete square, so the square can be taken out. The only complete square among the factors of 20 is 4, so:

$x = -2 \pm {\sqrt{(4)(5)} \over 2}$

$x = -2 \pm {2 \sqrt{5} \over 2}$

$2 \over 2$ cancels, so ...

$x = -2 \pm \sqrt{5}$

hope that helps.

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  • $\begingroup$ I asked if my simplification was correct which it appears it was. I was curious why the textbook did not simplify it further. $\endgroup$ – dagda1 Aug 11 '15 at 15:26
  • $\begingroup$ Sorry, I totally misread your question. I thought the simpler answer was what the textbook had. Yes, your simplification is totally correct, and good. When I read "This is the answer the textbook that I am using gives ..." I read it as "This [that follows] is the answer the textbook that I am using gives" ... my bad. $\endgroup$ – Daniel Goldfarb Aug 11 '15 at 18:16
  • $\begingroup$ it is really, really interesting that only one person answered the question I asked and I can tell that everybody who answered is pretty smart. I'm pretty sure I am guilty of this on stackoverflow :). $\endgroup$ – dagda1 Aug 11 '15 at 19:24

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