How many solutions would a quadratic equation have if the discriminant is 12?

Question

Adelyn French · Accepted Answer

A quadratic equation has two distinct real roots if the discriminant is greater than or equal to 12.
Due to the fact that 12 is not a perfect square, if the quadratic's coefficients are integers or otherwise rational, both of its roots must be irrational.
To see that the condition on the coefficients being rational is necessary for both roots to be irrational, consider the equation:
\(\displaystyle{x}^{{2}}-{\left({2}+\sqrt{{{12}}}\right)}{x}+{\left({1}+\sqrt{{{12}}}\right)}={0}\)
This is in the form \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}\) with a=1, \(\displaystyle{b}=-{2}-\sqrt{{{12}}}\), \(\displaystyle{c}={1}+\sqrt{{{12}}}\)
It has discriminant:
\(\displaystyle\Delta={b}^{{2}}-{4}{a}{c}\)
\(\displaystyle{\color{white}{\Delta}}={\left(-{2}-\sqrt{{{12}}}\right)}^{{2}}-{4}{\left({1}\right)}{\left({1}+\sqrt{{{12}}}\right)}\)
\(\displaystyle{\color{white}{\Delta}}={\color{red}{\cancel{{{\color{black}{{4}}}}}}}+{\color{red}{\cancel{{{\color{black}{{4}\sqrt{{{12}}}}}}}}}+{\left(\sqrt{{{12}}}\right)}^{{2}}-{\color{red}{\cancel{{{\color{black}{{4}}}}}}}-{\color{red}{\cancel{{{\color{black}{{4}\sqrt{{{12}}}}}}}}}\)
\(\displaystyle{\color{white}{\Delta}}={12}\)
We find:
\(\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\)
\(\displaystyle{\color{white}{{x}}}=\frac{{-{b}\pm\sqrt{{\Delta}}}}{{{2}{a}}}\)
\(\displaystyle{\color{white}{{x}}}=\frac{{{\left({2}+\sqrt{{{12}}}\right)}\pm\sqrt{{{12}}}}}{{2}}\)
That is:
\(\displaystyle{x}={1}\ \text{ }\ #{\quad\text{or}\quad}#\ \text{ }\ {x}={1}+\sqrt{{{12}}}\)

How many solutions would a quadratic equation have if the discriminant is 12?

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