Late post, but in this blog, I will talk about what I’ve learned last week (week 7) in Precalc 11!

To briefly go over it, one of the things I’ve learned is about the discriminant. The discriminant is a part of the quadratic formula used to find ** how many **“roots,” “solutions,” “zeros,” or what we usually call as “x’s” in our formula.

Before we go over that, let’s talk about the quadratic formula quickly.

this is the formula used *to find* the “x’s” of a quadratic equation. This is usually used when you can’t “factor” your equation.

Anyways, the “discriminant” that I was talking about is the root part of the equation or “b^{2 }– 4ac”

What does it do?

Like I’ve said, it’s used to find if there are any roots or solutions in your equation and how many there is(are).

So how exactly do we determine how many “roots” there is?

**NOTE: Examples used are easier to factor just for easier explanation, but typically you use the quadratic formula if there is no way of factoring it.**

**Situation one: b ^{2 }– 4ac > 0**

When the discriminant is **more than zero **or **positive**, then equation ax^{2} + bx + c = 0 (quadratic) has **two real, unequal roots**.

i.e. **x ^{2} + 5x + 4 = 0**

a = 1; b = 5; c = 4

b^{2} – 4ac > 0

5^{2} – 4(1)(4) > 0

25 – 16 > 0

9 > 0

So this means that x^{2} + 5x + 4 = 0 has two unequal, and real roots.

Notice that if you’d factor it, it will be (x + 4) (x + 1), in which your roots are

x = -4 and x = -1, that’s why you’d have two real, unequal roots.

**Situation two: b ^{2 }– 4ac = 0**

When the discriminant is **zero, **then there will be **two real and equal roots**. Technically, there is only “one” solution because both x’s have the same value. This usually occurs in perfect square trinomials.

i.e. x^{2} + 4x + 4 = 0

a = 1; b = 4; c = 4

b^{2} – 4ac = 0

4^{2} – 4(1)(4) = 0

16 – 16 = 0

0 = 0

This means that the equation x^{2} + 4x + 4 = 0 only has one real root. But notice how you’d know it immediately if you factor a perfect square trinomial.

x^{2} + 4x + 4 = 0

(x + 2) (x + 2) = 0

**x = -2; x= – 2. They just have the same value for both solutions, so technically, even though there are two solutions, we can say that it only has one.**

**Situation three: b ^{2 }– 4ac < 0**

When the discriminant is **less than zero **or **negative**, then the equation has **no roots or solutions.**

Why? Well, think about it. The quadratic formula is like this: r

And we use the discriminant part, which is under a square root sign to figure out the roots.

So, if your discriminant is negative, then it will be an error because you can’t calculate the square root of a negative number.