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Given that
$a^2 - a + 1 = 0$,

my book says:
Therefor $a = \frac{1}{2}(1\pm\sqrt{1+4}).$

I have forgotten all the theory behind this.

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4 Answers 4

up vote 2 down vote accepted

Hint

Quadratic Formula....

The solutions of a quadratic equation, $ax^2+bx+c=0$ with $a \neq 0$, is given bu the quadratic formula: $$x=\frac{-b+\sqrt{b^2-4ac}}{2a}$$

Remark: The quantity within the square root, $b^2-4ac$ is called the discriminant of the equation and is denoted by $\Delta=b^2-4ac$. The sign of $\Delta$ helps differentiate the nature of roots of the equation.

Related Reading on this site:

  1. Why can ALL quadratic equations be solved by the quadratic formula?
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There must be a typo in your post or in the book. The equation $a^2-a+1=0$ has no real solutions. The (complex, non-real) solutions, by the Quadratic Formula, are $$a=\frac{1\pm\sqrt{-3}}{2}.$$

Remark: If the original equation was $a^2-a-1=0$, then the roots would indeed be as given. The equation $x^2-x-1=0$ comes up fairly often, for instance in work connected with Fibonacci numbers. The positive root $\frac{1+\sqrt{5}}{2}$ is sometimes called the golden number, and even has two standard symbols for it, $\phi$ and $\tau$.

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Here's the way I like to think of quadratic equations where the $x^2$ coefficient is 1. Let's say the equation has roots $r_1$ and $r_2$. Then our equation is

$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2=0$

So as you can see, the sum of the roots is equal to $-1$ times the x coefficient and the constant term is the product of the roots. If our roots are $\frac12\pm\frac{\sqrt5}2$, the sum of the roots is $2(\frac12)=1$ and the product of the roots is

$(\frac12+\frac{\sqrt5}2)(\frac12-\frac{\sqrt5}2)=(\frac12)^2-(\frac{\sqrt5}2)^2=\frac14-\frac54=-1$.

So the equation for these roots should be $a^2-a-1=0$. So what should the roots be for $a^2-a+1=0$? Well, the sum of the roots is $1$, so the average of the 2 roots is $\frac12$. So let's say our roots are $\frac12\pm d$. We want the product of the roots to be $1$. So we have

$(\frac12+d)(\frac12-d)=\frac14-d^2=1$

$d^2=-\frac34,d=\pm\frac{i\sqrt3}2$

So our roots are $\frac12\pm\frac{i\sqrt3}2$

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$$\begin{eqnarray*} a^2 - a +1&=& a^2 -a +\frac{1}{4} + \frac{3}{4} \\ &=& a^2 - 2\left(\frac{a}{2}\right) + \frac{1}{4} + \frac{3}{4} \\ &=& \left(a - \frac{1}{2}\right)^2 + \frac{3}{4}\\ \end{eqnarray*}$$

so if $a^2 - a + 1 =0$, this means that

$$\begin{eqnarray*} \left(a - \frac{1}{2}\right)^2 + \frac{3}{4} &=& 0 \\ \implies \left(a - \frac{1}{2}\right)^2 &=& \frac{-3}{4}\\ \implies a &=& \frac{1}{2} \pm \sqrt{\frac{-3}{4}}. \end{eqnarray*}$$

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