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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Home -> Solved problems -> Discriminant of quadratic equation