Question

Solve x^2 - 3x + 1 = 0 algebraically

Answer 1

I used formula:
\[ax^{2}+bx+c = a(x-\frac{-b-\sqrt{b^{2}-4ac}}{2a})(x-\frac{-b+\sqrt{b^2{}-4ac}}{2a}) = 0\]
Which is pretty cheap and lame.

Is there a better/simpler way to solve it?

Answer 2

What do you mean "better and simpler"

If you're looking for another way to solve it just complete the square, if you're looking for a fancy way, then I don't know.

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude
What do you mean "better and simpler"
\(\color{#0cbb34}{\text{End of Quote}}\)
Like usually I can rewrite it in (x-b)(x-d) = 0 by looking at it but I couldn't for this one.

I will try completing the square.

\[x^{2}-3x+(\frac{3}{2})^{2}=-1+(\frac{3}{2})^{2}\]
\[(x-\frac{3}{2})^2 = \frac{4}{4}+\frac{9}{4}\]
\[x-\frac{3}{2} = \pm \sqrt{\frac{13}{4}}\]
\[x = \frac{3}{2}+\sqrt{\frac{13}{4}},x=\frac{3}{2}-\sqrt{\frac{13}{4}}\]
lol it works