$\bf 2=x^2+5x\implies 0=x^2+5x-2\\\\\\\qquad \qquad \qquad \textit{discriminant of a quadratic}\\\\\\0=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+5}x\stackrel{\stackrel{c}{\downarrow }}{-2}~~~~~~~~\stackrel{discriminant}{b^2-4ac}=\begin{cases}0&\textit{one solution}\\positive&\textit{two solutions}\\negative&\textit{no solution}\end{cases}\\\\\\(5)^2-4(1)(-2)\implies 25+8\implies 33$

so we have a 33, namely two real solutions for that quadratic.

usually that number goes into a √, if you have covered the quadratic formula, you'd see it there, namely that'd be equivalent to √(33), now 33 is a prime number, and √(33) is yields an irrational value, specifically because a prime number is indivisible other than by itself or 1.

so 33 can only afford us two real irrational roots.

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3 years ago