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I was working on some quadratic equations and wondered if I could try and solve an equation without using the formula. Could you tell me if my method is correct? I tried solving it this way to improve my command over the mathematical language and to see if I've understood the concept correctly. What I did was take an equation who's x^2 coefficient was > 1 and convert it into the vertex form to further solve it.

The way I went about it was:

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    $\begingroup$ You made an error in the second line: it should be $\frac{12}{9}$ instead of $\frac{20}{9}$. Otherwise it looks ok. $\endgroup$ – Loreno Heer Mar 4 '17 at 15:42
  • $\begingroup$ Great! Thanks a lot :) I had another question. In case if c has a positive value rather than a negative would it mean that equation has no real roots? So for example, if the equation was 3x^2+4x+1 would it mean that the equation has no real roots because then I'd have to do square root of -1.. $\endgroup$ – user401747 Mar 4 '17 at 15:45
  • $\begingroup$ Completing the square and solving for x is typically how a class proves the formula. So what you did is equivalent. Yes, if you have to take the square root of a negative number, the roots aren't real. $\endgroup$ – Kaynex Mar 4 '17 at 15:50
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we have to solve $$3x^2+4x-1=0$$ dividing by $3$ we get $$x^2+\frac{4}{3}x-\frac{1}{3}=0$$ then we have $$x^2+2\cdot \frac{2}{3}x+\frac{4}{9}-\frac{1}{3}-\frac{4}{9}=0$$ the first three Terms are a complete square $$\left(x+\frac{2}{3}\right)^2=\frac{1}{3}+\frac{4}{9}$$ the right side is given by $$\frac{1}{3}+\frac{4}{9}=\frac{7}{9}$$ thus we have $$|x+\frac{2}{3}|=\frac{\sqrt{7}}{3}$$ from here we get $$x=-\frac{2}{3}+\frac{\sqrt{7}}{3}$$ $$x=-\frac{2}{3}-\frac{\sqrt{7}}{3}$$

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  1. You have missed one root when taking the square root from both sides. However, you've recover it on the next row.
  2. You can check your roots on wolframalpha. As Loreno Heer stated in the comment to the question, there is a mistake in your derivation which then resulted in wrong roots.
  3. In fact you have just reproduced the flow of derivation of the descriminant of quadratic equations. If you want to avoid these computations, which could be could be quite "extensive" on the exams, you could look at Vieta's formulas for quadratic equation. It is especially effective when you have $a = 1$ in the equation $$P(x) = ax^{2}+bx+c=0$$ So then you have $$x_{1}+x_{2}=-b,\quad x_{1}x_{2}=c,$$ where $x_{1,2}$ are your roots. So, you can pick integer divisors of $c$ and check if one is a root of $P(x) = 0$. In the school practice, it is quite often that after 1-2 iterations you finally find the root. Moreover, you do not need to mess up with the square rooting of anything, so a mistake is less likely. Once one root $r_1$ is found, the other is $r_2 = -b - r_1$. But, of course, it works only for quadratic equations with integer roots. Otherwise you sould better consider the discriminant formula.
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Using this approach you can derive a formula for solving quadratic equations. Let $a x^2 + b x - c$ be a polynomial with ($a \neq 0$), then:

$$a x^2 + b x = -c \iff a(x^2 + \frac{b}{a}x) = c$$ $$\iff a(x+\frac{b}{a}x +\frac{b}{2a}) = c + \frac{b}{2a}$$ $$\iff a(x+\frac{b}{2a})^2 = c + \frac{b}{2a}$$ $$\iff \pm\sqrt{a}(x+\frac{b}{2a}) = \sqrt{c+\frac{b}{2a}}$$ $$\iff x = \frac{\pm\sqrt{c+\frac{b}{2a}}}{\sqrt{a}} - \frac{b}{2a} $$

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