+6251 \(\mbox{If you have a quadratic polynomial }a x^2 + b x + c \mbox{ the roots of this are given by}\\ r_{1,2}=\dfrac{-b\pm\sqrt{b^2-4 a c}}{2a} \\ \mbox{The argument of the square root in this expression }(b^2 -4 a c) \mbox{ is called the discriminant.}\)
\(\mbox{It's called this because it's value discriminates between real and complex roots.}\\ \mbox{if it is non-negative then the roots of the quadratic will be real. If it is negative} \\ \mbox{they will be a complex conjugate pair.}\)
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Go to this website and WATCH the video:https://www.khanacademy.org/test-prep/fr-twelveth-grade-math/Nombres-complexes/La-formule-des-racines-d'un-trin%C3%B4me/v/discriminant-of-quadratic-equations.
+6251 \(\mbox{If you have a quadratic polynomial }a x^2 + b x + c \mbox{ the roots of this are given by}\\ r_{1,2}=\dfrac{-b\pm\sqrt{b^2-4 a c}}{2a} \\ \mbox{The argument of the square root in this expression }(b^2 -4 a c) \mbox{ is called the discriminant.}\)
\(\mbox{It's called this because it's value discriminates between real and complex roots.}\\ \mbox{if it is non-negative then the roots of the quadratic will be real. If it is negative} \\ \mbox{they will be a complex conjugate pair.}\)
+118667 Rom is correct of course but it also discriminates between other tytpes of roots as well.
The symbol for the discriminant is a triangle.
so
\(if \;\;ax^2+bx+c=0\qquad then\qquad x=\frac{-b\pm\sqrt{b^2-4ac}}{2}\\ The \;discriminant \;is\;b^2-4ac\\ \triangle=b^2-4ac \)
IF the discriminant = zero there will be one real root. (sometimes called a double root)
IF the discriminant > 0 then there will be 2 real roots.
If the discriminant < 0 there there will be no real roots. If you have not done complex numbers yet then you would probably say that there were NO roots. This is because you cannot take the square root of a negative number.
ROM already said all this.
BUT also
If the descriminant is a perfect square then the roots will be rational
If the descriminant is NOT a perfect square then the roots will be irrational
