If b²-4ac > 0
There are 2 real roots
If $b^2-4ac>0$, then the quadratic equation $ax^2+bx+c=0$ has two real roots.
To understand this, let’s start with the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
The quadratic formula tells us that the roots of a quadratic equation depend on the discriminant, which is the expression under the square root sign in the formula: $b^2-4ac$.
If the discriminant is positive, then we will have real and distinct roots. This means that the quadratic equation will have two different solutions, and both solutions will be real numbers.
For example, consider the quadratic equation $x^2-5x+6=0$. Here, we have $a=1$, $b=-5$, and $c=6$, so the discriminant is:
$$b^2-4ac = (-5)^2 – 4(1)(6) = 1$$
Since the discriminant is positive (and not equal to zero), we know that the quadratic equation has two distinct roots. Using the quadratic formula, we can find those roots:
$$x = \frac{5 \pm \sqrt{1}}{2}$$
So the roots are $x=2$ and $x=3$, both of which are real numbers.
In summary, if $b^2-4ac>0$, then the quadratic equation has two real roots.
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