## Find out what is a discriminant of a quadratic equation.

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### Solution

Let’s take the general form of a quadratic equation:
$ax^{2}+bx+c=0\;\;\;\;\;\;\; a\neq0$
\begin{aligned} &x^{2}+\frac{b}{a} x+\frac{c}{a}=0 \\\\ &x^{2}+\frac{b}{a} x+\left(\frac{b}{2a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}=0 \\\\ &\left(x+\frac{b}{2 a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}=0 \\\\ &\left(x+\frac{b}{2 a}\right)^{2}=\left(\frac{b}{2 a}\right)^{2}-\frac{c}{a}=\frac{b^{2}}{4 a^{2}}-\frac{c}{a} \\\\ &\left(x+\frac{b}{2 a}\right)^{2}=\frac{b^{2}-4 a c}{4 a^{2}} \end{aligned}
If $$b^{2}-4ac<0$$ then there are no real roots for the quadratic equation.
If $$b^{2}-4ac=0$$ then the quadratic equation has two real, identical roots:
$\begin{gathered} x_{1}=x_{2}=-\frac{b}{2 a} \\\\ a x^{2}+b x+c=a\left(x+\frac{b}{2 a}\right)^{2} \end{gathered}$
If $$b^{2}-4ac>0$$ then the quadratic equation has two real, distinct roots:
\begin{aligned} &x_{1}=\frac{-b-\sqrt{b^{2}-4ac}}{2a} \\\\ &x_{2}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\\\ &ax^{2}+bx+c=a\left(x-\frac{-b-\sqrt{b^{2}-4ac}}{2 a}\right)\left(x-\frac{-b+\sqrt{b^{2}-4ac}}{2a}\right) \end{aligned}
Home -> Solved problems -> Discriminant of quadratic equation

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