I don't understand why this doesn't work:
Problem: $$x^{2}+x-2=0$$ Solution: $$x^{2}+x=2$$ $$x+x=\sqrt{2}$$ $$2x≈1.4$$ $$x≈1.4/2$$ $$x≈0.7$$ Checking: (Note: I made the calculation with more decimals than the amount I wrote here; that explain $0.7^{2}=0.5$) $$0.7^{2}+0.7-2≈0.5+0.7-2≈-0.79$$ This method doesn't work...it gives $-0.79$ and it must give 0.

Instead I must do:

Good Solution: $$\frac{-b±\sqrt{b^{2}-4ac}}{2a}=0$$ $$\frac{-1±\sqrt{1^{2}-4\times1\times(-2)}}{2\times 1}=0$$ $$\frac{-1±\sqrt{1+8}}{2}=0$$ $$\frac{-1±\sqrt{9}}{2}=0$$ $$\frac{-1±3}{2}=0$$ $$x_{1}=\frac{-1+3}{2}=\frac{2}{2}=1$$ $$x_{2}=\frac{-1-3}{2}=\frac{-4}{2}=-2$$ Checking: $$1^{2}+1-2=1+1-2=0$$ $$(-2)^{2}+(-2)-2=4-2-2=0$$ And that gives a good result.

  1. I don't understand why my first attempt doesn't work.
  2. Also, I don't understand why it has 2 possible results, $1$ and $-2$.
  • 3
    $\begingroup$ About 2: why is it strange that there are two solutions? Equations are welcome to have more than one solution. Remember that solving your equation is equivalent to asking "Which numbers fulfill the property that if you square it, then add the original number to the result, you get $2$" There is no universal principle that says that only one number can fulfill a given property. $\endgroup$ – Arthur Nov 23 '17 at 17:27
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    $\begingroup$ To see why there are two solutions, observe that you can factor your expression: $$x^2+x-2 = (x+2)(x-1)$$ So, if $(x+2)(x-1) = 0$, then either $x+2=0$ or $x-1=0$. $\endgroup$ – Théophile Nov 23 '17 at 17:27
  • $\begingroup$ @Arthur you have right. It only matter that one of the two brackets has 0, beacuse $anything \times 0 = 0$.Thanks. $\endgroup$ – Ender Look Nov 23 '17 at 17:37
  • $\begingroup$ @Ender I had edited and corrected the quadratic formula, because it was incorrect. You can check it any where. There must be $b^2$ under the square root, instead of $a^2$ $\endgroup$ – Jaideep Khare Nov 23 '17 at 18:04
  • $\begingroup$ @JaideepKhare, ups, I accidentaly wrote that wrong, thanks. $\endgroup$ – Ender Look Nov 23 '17 at 19:17

This is wrong :

$$x^2 + x = 2 \Rightarrow x + x = \sqrt{2}$$

The square root applies to the whole term :

$$x^2+x=2 \Rightarrow \sqrt{x^2+x} = \sqrt2$$

if $x^2 + x \geq 0$.

This is why your attempt is mistaken.

The equation $x^2 + x - 2=0$ has two roots (not two possible results, this is mistaken as a phrase), because of the square power. Explaining why this form means $2$ solutions, can be done in a lot of ways.

Graphically and geometrically, an equation of the form : $ax^2 + bx + c = 0$ describes a parabola, which due to its geometric form, if it has one solution, it needs to have another one, thus two roots, either on the real plane, or in the complex plane, based on the sign of its discriminant.

Alternatively, you can observe two numbers that satisfy the equation, by factorization : $(x-1)(x+2)=0$, which means that either for $x=1$ or $x=-2$ the equation/expression is satisfied.

Equations of the form $ax^2 + bx + c = 0$, which is the ones your post is interested in, always have a straight-forward solution and are of the most common in entry mathematics.

You can read up on some more stuff for the quadratic equation here.

  • 2
    $\begingroup$ Thanks for your answer, I didn't understand that about the parabola but the alternative explanation (by factorization) was very easy to understand and has satisfied my doubts, thank. $\endgroup$ – Ender Look Nov 23 '17 at 19:31
  • $\begingroup$ Think about the U-shape of a parabola. Then if it intersects the $x'x$ axis at even one point, then wouldn't it intersect it at one more ? If the $x'x$ axis is tangent at the global minima or maxima of the U-shape, then it's the case of the discriminant being equal to 0, which gives you two equal solutions, for example : $(x-2)^2=0$ which has one solution but of double multiplicity, $x=2$. $\endgroup$ – Rebellos Nov 23 '17 at 19:52
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    $\begingroup$ Unfortunately, I don't think the OP know what a parabola is, or what the graph of a function is. $\endgroup$ – user202729 Nov 24 '17 at 11:39

This doesn't work because the step $$x^2+x=2 \implies x+x=\sqrt 2$$ is wrong.

You cannot take root only on one term in LHS, you will have to take square root of complete LHS. That is $\sqrt{x^2+x}$.

I think an example will make it easy for you to understand.

What you did : $$16+9=25 \implies \sqrt{16}+9=\sqrt{25} \implies 4+9=5$$ and this is definitely wrong.

  • $\begingroup$ Sorry, english isn't my native language, what means LHS? $\endgroup$ – Ender Look Nov 23 '17 at 19:40
  • $\begingroup$ @EnderLook it means Left Hand Side (of the equality/inequality) also, RHS means Right Hand Side. Although your native language is Spanish, but you will come across this abbrevation many times, as it is very commonly in all field of mathematics. For example : In the equation $$x=2$$ $x$ is on LHS and $2$ is on RHS. $\endgroup$ – Jaideep Khare Nov 23 '17 at 19:43
  • $\begingroup$ @Ender Look. Imagine instead of zero you have 'y'. As you change x, look what happens to y. Y will be vertical position on a graph. X will be horizontsl. The U shape will cross the x axis twice. That is your zero. $\endgroup$ – Sentinel Nov 23 '17 at 22:01
  • $\begingroup$ @Sentinel I think you need to write this in the comment section of accepted answer. $\endgroup$ – Jaideep Khare Nov 23 '17 at 22:02
  • $\begingroup$ @jaideep khare. Ah yes, you are right. $\endgroup$ – Sentinel Nov 23 '17 at 22:08

Your step from $x^2+x=2$ to $x+x=\sqrt 2$ is not correct. First you cannot take the square root of just one term on one side of the equation. The proper expression would be $\sqrt{x^2+x}=\sqrt 2$ which is more complicated, not less, so probably a step in the wrong direction. Second, taking a square root introduces a sign ambiguity. If you had $x^2=4$ and took the square root you need to get $x=\pm 2$ because $(-2)^2=4$ as well as $2^2=4$.

  • $\begingroup$ That means that for each square (or pair) root in an quation I would have two aswers (one in positive and the another in negative)? $\endgroup$ – Ender Look Nov 23 '17 at 19:35

Why are there two results? Well $(-a)^2 = (a)^2$ and though there is a convention that the principal (non-negative) value of the square root is used when we take the square root, when we operate by squaring a number, there are two numbers which give the same result - one is positive and the other negative.

So the template equation $$x^2=a^2$$has two solutions $x=\pm a$ unless we happen to have $a=0$ when there is one solution.

The template equation can also be written in the form $x^2-a^2=(x+a)(x-a)=0$.

To solve a quadratic we want to put it in template form. This is often easiest if we multiply the equation $ax^2+bx+c=0$ by $4a$ before doing any further work.

Here we multiply by $4$ to get $$4x^2+4x-8=0$$

To get template form we then "complete the square" (in the general case this gives a term $(2ax+b)^2$). Here we recognise $4x^2+4x$ as the first two terms of the square $$(2x+1)^2=4x^2+4x+1$$ so we see that $$(2x+1)^2=9=3^2$$

And we now have a template form and we can deduce $2x+1=\pm 3$ - the two possibilities coming from the two signs of the square roots we get from the template.

Notice how, in using the template, we needed to deal with both the terms involving $x$, so that we could take the square root - this is what we mean by "completing the square". Since we don't know what $x$ is, if we don't do this, we are left with terms we can't evaluate.

  • 1
    $\begingroup$ "There is a convention that the principal (non-negative) value of the square root is used when we take the square root" - this is rather imprecise (IMHO dangerously so if you are talking to a beginner). It would be better to say that the non-negative value of the square root is taken when we use the $\sqrt{\phantom1}$ symbol. $\endgroup$ – David Nov 23 '17 at 23:26
  • $\begingroup$ @David Would the same not apply to exponent $1/2$? What I was trying to say is that there are two values which square to some number $a$, one of which is conventionally taken as the square root. (Of course that does not apply to the quaternions, but are these in view?) What is the danger here in the way I have expressed things? $\endgroup$ – Mark Bennet Nov 23 '17 at 23:34
  • $\begingroup$ Hi Mark, I should make it clear that I think your answer as a whole is fine, it's just the first part that concerns me. The distinction is that by convention, $\sqrt4$ is $2$, but the solution of $x^2=4$ is $x=\pm2$. It's not actually a question about solving equations, but about the use of notation. As far as I know the same convention applies to $()^{1/2}$. $\endgroup$ – David Nov 23 '17 at 23:44

$x^{2}+x=2$ does not imply $x+x=\sqrt{2}$

In fact $x^{2}+x$ does not have a simpler square-root, but $x^2+x+\frac14$ does because it is $\left(x+\frac12\right)^2$

So let's try that:

$$x^{2}+x-2=0$$ $$x^{2}+x+\frac14=\frac94$$ $$x+\frac12=\frac32 \text{ or } -\frac32$$ $$x=1 \text{ or } -2$$


Here is a step by step method:

\begin{align} x^2 + x - 2 = 0 \\ x^2 + 2\frac{1}{2}x + \frac{1}{4} - \frac{1}{4} -2 = 0 &&&\text{Completing the square} \\ (x + \frac{1}{2})^2 - \frac{1}{4} - 2 = 0 &&& a^2+2ab+b^2 = (a+b)^2 \\ (x + \frac{1}{2})^2 - \sqrt{\frac{1}{4} + 2}^2 = 0 \\ \left(x+\frac{1}{2}-\sqrt{\frac{1}{4} + 2}\right)\left(x+\frac{1}{2}+\sqrt{\frac{1}{4} + 2}\right) = 0 &&& a^2-b^2 = (a-b)(a+b) \end{align}

We have found our solution $x^\pm = -\frac{1}{2}\pm\sqrt{\frac{1}{4}+2}$. These are exactly the same as the solutions of the form $\displaystyle x^\pm = -\frac{b\pm\sqrt{b²-4ac}}{2a}$ that you want. Note that $\sqrt{\frac{1}{4}+2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$, so we can simplify: \begin{align} \left(x+\frac{1}{2}-\frac{3}{2}\right)\left(x+\frac{1}{2}+\frac{3}{2}\right) = 0 \\ \left(x-\frac{2}{2}\right)\left(x+\frac{4}{2}\right) = 0 \\ \left(x-1\right)\left(x+2\right) = 0 \end{align} Yielding the solutions $x_1=1$ and $x_2 = -2$.

  • 1
    $\begingroup$ The way I completed the square was after your third line moving the non-x terms to the right hand side and taking the square root of both sides (adding a +- onto the right hand side) and then taking half off both sides. It amounts to the same thing but feels simpler because you aren't relying on the difference of two squares formula. $\endgroup$ – Chris Nov 24 '17 at 12:45
  • $\begingroup$ @Chris I feel the $\pm$ symbol is slightly deceiving, as it actually a short way to describe two equations. You also have to explain why only one side gets the $\pm$ symbol, and it may cause people to believe that the square root can be both positive and negative. So there's many subtleties one should be made aware of in that case. The difference between two squares is no more difficult than completing the square. $\endgroup$ – Frank Vel Nov 24 '17 at 13:52
  • $\begingroup$ Yeah, I see your point. I guess my comment was not so much to suggest you change how you did it as to highlight the fact that this is taught differently in different places. Though the ± is no more ambiguous here as it is in the quadratic formula and you could do it on both sides if you wanted to and get the same results. $\endgroup$ – Chris Nov 24 '17 at 14:25

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