Discriminant to Find Nature of Roots

Have you ever wondered what type of solutions a quadratic equation will have without actually finding the solutions?

The discriminant does just that! It is the part of the quadratic formula that appears directly under the radical sign. Take a look at the photo shown below:

Based on what the value of the discriminant is we can tell what types of solutions we will have.

For instance, imagine the discriminant is a negative number. In this scenario, we will have a negative number under the radical sign. This would indicate 2 imaginary solutions.

Now let's imagine another scenario. If the discriminant equals zero we will now have zero under the radical sign. Well the square root of zero is zero so we would essentially be adding zero to negative b in the numerator. Because adding or subtracting zero yields the same response, having a discriminant value of zero would indicate only one real solution. This is often called a "double solution" or "two equal solutions"

The last scenario is when the discriminant is positive. In this case, we will have two real solutions. The solutions cannot be imaginary because the number under the radical sign is not negative. If the positive discriminant value is a perfect square, we will have two rational solutions. If the discriminant value is a non perfect square, we will have two irrational solutions.

This chart below should summarize the information about the discriminant.

Now that we know what quadratics are, a pressing question arises. Just as we were able to find solutions to linear equations like $2x + 3 = 5$, can we find the solutions to a quadratic like $5x^2 - 12x + 5 = 1$?

Let's start simple.

Think about the solutions to $x^2 - 9 = 0$.

We know that the solutions are $x=3$ and $x=-3$. The equation $x^2 - a = 0$ always has two solutions for $x$: $\sqrt{a}$ and $-\sqrt{a}$. The only exception is $a = 0$.

Now, think about what the solutions are to $(x-1)^2 = 9$.

We have two cases: $x - 1 = 3$ or $x - 1 = -3$. This gives the solutions $-2$ and $4$.

Diving In

Think about the solutions to $x^2 + 12x + 20 = 48$.

We can start by trying to make a perfect square using $x$. Note that $x^2 + 12x + 36 = (x+6)^2$; thus, if we add $16$ to both sides of our equation, we get $(x+6)^2 = 64$. That's great -- we know how to solve that! Since $64 = 8^2$, we have $x+6 = 8$ or $x+6 = -8$. In the first case we get $x = 2$, and in the second we get $x = -14$. Hence those are our two solutions.

Let's move things up a notch.

Let's get rid of that $3$ in front of the $x^2$ by dividing both sides of the equation by it. We obtain $x^2 - 4x + \frac{25}{3} = \frac{16}{3}$. Now in order to make a perfect square, we want the left-hand side to be $x^2 - 4x + 4$; hence, we subtract $\frac{13}{3}$ from both sides to get $x^2 - 4x + 4 = 1$. And since $x^2 - 4x + 4 = (x-2)^2$, we now have two cases: $x-2 = 1$ or $x-2 = -1$. They give the two solutions $x = 3$ and $x = 1$.

Sweet! Now we're ready to tackle the general form of a quadratic equation. If we have the equation $ax^2 + bx + c = 0$, how can we solve for $x$ in terms of $a$, $b$, and $c$?

Just like the last problem, we start by getting rid of the $a$ in front of the $x^2$. So after we divide, the equation becomes $$x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0.$$ Now we want to form a perfect square. The ideal square is $$\left(x + \dfrac{b}{2a}\right)^2 = x^2 + \dfrac{b}{a}x + \dfrac{b^2}{4a^2},$$ so we subtract $\frac{c}{a}$ and add $\frac{b^2}{4a^2}$ to both sides of our equation. That leaves us with $$\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a}.$$

All right, we're getting close! We can combine the two fractions on the right-hand side by making a common denominator: $$\dfrac{b^2}{4a^2} - \dfrac{c}{a} = \dfrac{b^2 - 4ac}{4a^2}.$$ Now we have $$\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2-4ac}{4a^2}.$$ Hence $$x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2-4ac}}{2a}.$$ Finally, moving over the $\frac{b}{2a}$ term, we conclude that

$$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

At last, we're done! This is known as the quadratic formula. It can be used to find the solutions to any quadratic equation. Learn to love it; it'll come up a lot in algebra for its usefulness.

In fact, general formulas (although far, far more complicated) also exist for cubic and quartic equations, whereas it is impossible to find one for 5th (or higher) degree equations. Though the proof of that last fact is far beyond the scope of this post :)

Let our equation be of the form $$x^2+bx+c=0$$ and let $x_1$ and $x_2$ be the roots of the quadratic.

Then $$x^2+bx+c=(x-x_1)(x-x_2)$$ $$=x^2-(x_1+x_2)x+x_1x_2$$ and therefore $x_1+x_2=-b$ and $x_1x_2=c$. Now, if we knew $d=x_1-x_2$, we could figure out $x_1=(-b+d)/2$ and $x_2=(-b-d)/2$.

The trouble is that our equation does not know which of the roots is $x_1$ and which one is $x_2$, so we can know $d$ only up to a sign, so we can know $d^2$. Let us try to figure it out in terms of $b$ and $c$. Can you do it?

Here comes a little magic: $$d^2=x_1^2-2x_1x_2+x_2^2$$ and $$b^2=x_1^2+2x_1x_2+x_2^2,$$ they differ just by $4x_1x_2=4c$. Ta-taam! Now we get $d^2=b^2-4c$ and finally we get our quadratic formula: $$x_{1,2}=(-b\pm \sqrt{b^2-4c})/2$$

What is the Quadratic Formula?

When given an equation in the form $ax^2 + bx + c$, usually the simplest way to solve for the value of $x$ is to factor the quadratic, set each factor equal to zero, and then solve for each factor. However, the quadratic can sometimes be too difficult to solve by factoring. At this point, you may want to throw in the towel, give up on math, and run far away. Never fear! You can count on the quadratic formula to find the solution for you!

When using the quadratic formula, you will need to have an $”a”, “b”,$ and $”c”$ term from $ax^2 + bx + c$, where $”a”, “b”,$ and $”c”$ are numbers. These are considered numerical coefficients of the quadratic equation that you are asked to solve.

You may be asking, “but what IS the quadratic formula?” Well, here it is, derived from the process of completing the square (which you can refresh on here!) $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

This beautiful equation will be engrained into your mind forever. I learned it five years ago, and I still remember it to this day!

Rules for the Quadratic Formula

1) In order for the quadratic formula to work, you must have the equation arranged so that $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ is set equal to $0$.

2) Make sure that you are careful not to drop the square root, or the $\pm$ before the square root. Because you have the $\pm$, this means you will have $2$ roots!

3) Our $a$ term can NEVER be $0$. This will cause the denominator of the quadratic formula to be $0$, and it will void the equation since you can never EVER divide by $0$. If you are given an equation in the form of $bx + c = 0$, you will have to solve for $x$ by subtracting both sides by $c$, then dividing both sides by $b$.

4) The $b^2$ under the square root means that you square ALL of $b$, including a negative sign if applicable. So, the $b^2$ term will never be negative!

5) Also, take note that you will multiply your $b$ by a negative sign at the beginning of the numerator.

All in all, make sure you take note of your signs! If you try to take shortcuts, or try to shorthand the problem, you will mess it up! Take the quadratic formula slow, and I promise you will become a pro!

Example of Quadratic Formula

Solve $x^2 + 7x – 4$

Since this is a quadratic equation in the form $ax^2 + bx + c$, we can say that our $a$ term is $1$, since the coefficient of $x^2$ is $1$ (even if you can’t see it!) Our $b$ term is $7$, and our $c$ term is $-4$. Make sure that you include the negative sign in your $c$ term, or you’ll get the wrong answer!

With $a = 1$, $b = 7$, and $c = -4$ we can plug these values into the quadratic formula to get: $$x=\frac{-7\pm\sqrt{7^2-4(1)(-4)}}{2(1)} = 0$$ We can simplify what is being multiplied, to get: $$x=\frac{-7\pm\sqrt{49 + 16}}{2} = 0$$

NOTE that under the square root, we now have a summation of the numbers rather than a subtraction, because our $c$ term is negative! (and we know that two negative numbers multiplied together give us a positive!)

We can further simplify this to: $$x=\frac{-7\pm\sqrt{65}}{2} = 0$$

Now, since $65$ is not a perfect square, and it does not contain a perfect square, we can leave our answer in this form!

Great! With some more practice, you'll be a quadratic equation expert!

What is a Complex Root?

Now that you have had some practice with the quadratic formula, its time to move on to complex roots. This can sound scary at first, but I promise that once you get it, it will be a piece of cake! However, if you want a review, you can click the respective links below.

For a review on the quadratic formula, click here!

For a review on complex numbers, click here!

A complex root is a solution to a quadratic equation that also contains an $i$ term, similar to a complex number. Complex roots occur when the $b^2 - 4ac$ term is a negative number, since you will then be taking the square root of a negative number. This will become more apparent with an example.

Example of Complex Roots

Find the roots: $x^2 -4x + 13 = 0$ Let's define our $a,b$ and $c$ terms. For this expression, our $a$ term is $1$, our $b$ term is $-4$, and our $c$ term is $13$. Great, let's plug these into the quadratic equation! $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}$$ $$x=\frac{4\pm\sqrt{16-52}}{2}$$ $$x=\frac{4\pm\sqrt{-36}}{2}$$

At this point, we can see that we have a negative term under the square root. No worries though, we know what we need to do. We know that $36$ is a perfect square, therefore $-36$ is a perfect square as well. Taking the square root in the problem, we will get: $$x=\frac{4\pm 6i}{2}$$

$4$ and $6$ are divisible by $2$, so we can further simplify this to: $$x = 2 \pm 3i$$ Therefore, we have our final answer! The roots to our problem are $2 + 3i$ and $2 - 3i$.

And there you have it! Complex roots aren't as complex as you thought, huh?

Keep practicing, and you'll master this just like you mastered the quadratic formula the first time!

The Discriminant

The discriminant of a quadratic equation tells you the nature of the roots. That means you can use the discriminant to find out whether a quadratic function has two solutions, one solution, or imaginary solutions.

The discriminant is found within the quadratic equation, the value under the radicand.

When we solve the discriminant, if our value is:

• $d>0$, the quadratic has $2$ real solutions.

• $d=0$, the quadratic has $1$ real solution.

• $d<0$, the quadratic has $1$ imaginary solution.

(Video) Using the Discriminant to Find How Many Roots

by FerranteMath

In this video by FerranteMath, you will learn how to determine the number of roots in a quadratic by using the discriminant.

Summary

The discriminant is the expression found under the radical in the quadratic formula.

$$b^2-4ac$$

For any standard form of a quadratic, we can use the discriminant to find the number of roots that solve the quadratic equation.

If the parabola crosses the $x$-axis one time, the parabola is basically sitting on the $x$-axis. That means there is one real root or solution. That means that the discriminant is equal to zero.

$$b^2-4ac=0$$

If the parabola crosses the $x$-axis twice, there are two real roots or solutions. That means the discriminant is greater than zero.

$$b^2-4ac>0$$

If the parabola does not cross the $x$-axis, then there are no real solutions, but there are two imaginary roots. That means that the discriminant is less than zero.

$$b^2-4ac<0$$