Question

fects the roots of a quadratic equation. provide examples to support your explanation.
answer here

Explanation:

The discriminant of a quadratic equation $ax^2+bx+c=0$ is $\Delta = b^2-4ac$, and it determines the nature of the roots:

1. If $\Delta > 0$: Two distinct real roots.
2. If $\Delta = 0$: One repeated real root (a double root).
3. If $\Delta < 0$: No real roots (two complex conjugate roots).

Example 1: $\Delta > 0$

Take $x^2 - 5x + 6 = 0$, where $a=1$, $b=-5$, $c=6$.
$\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0$.
Roots: $x = \frac{5\pm\sqrt{1}}{2} = 3$ and $x=2$ (two distinct real roots).

Example 2: $\Delta = 0$

Take $x^2 - 4x + 4 = 0$, where $a=1$, $b=-4$, $c=4$.
$\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0$.
Root: $x = \frac{4\pm\sqrt{0}}{2} = 2$ (one repeated real root).

Example 3: $\Delta < 0$

Take $x^2 + 2x + 5 = 0$, where $a=1$, $b=2$, $c=5$.
$\Delta = (2)^2 - 4(1)(5) = 4 - 20 = -16 < 0$.
Roots: $x = \frac{-2\pm\sqrt{-16}}{2} = -1\pm2i$ (two complex conjugate roots, no real roots).

Answer:

The discriminant $\Delta = b^2-4ac$ of $ax^2+bx+c=0$ dictates root nature:

1. $\Delta > 0$: Two distinct real roots (e.g., $x^2-5x+6=0$ has roots 2 and 3)
2. $\Delta = 0$: One repeated real root (e.g., $x^2-4x+4=0$ has root 2)
3. $\Delta < 0$: No real roots, two complex roots (e.g., $x^2+2x+5=0$ has roots $-1\pm2i$)