Number of Roots of a Quadratic

The easiest way to think about a quadratic equation with no real solutions is to first visualize the graph.

First, look at this graph:

Using a simple process of elimination, we know that this equation has two real and non-equal roots because:

1. It passes through the $x$-axis twice. This indicates that it can be factored (if you don't believe me, try it yourself); in other words, you can solve this function and derive two $x$ values, which are the two $x$ values that this function passes through on the graph.
2. Its vertex is not on the $x$-axis. If it was, this would mean that the graph touches the $x$-axis at only one point. Since it is still a quadratic equation with an $x^2$ term, this means that it would have only one unique root. For example, a quadratic equation with only one unique real root would look like this:

This graph touches the $x$-axis at $x = -1$, which means its equation is $(x+1)(x+1)$ or $x^2 + 2x + 1$.

Finally, that leaves one more type of quadratic graph:

This graph does not touch the $x$-axis at any point, which means that it has no roots on the $x$-axis, and therefore no real roots. The roots are complex and can be found using the quadratic equation.

If you are not given the graph of the quadratic equation, you can use the discriminant to determine whether or not the quadratic has real roots.

The discriminant is the value known as $\sqrt{b^2 - 4ac}$, where $a$, $b$, and $c$ are the values in the quadratic equation $ax^2 + bx + c$. It can be used to 'discriminate' what types of roots a quadratic has.

If the discriminant is negative: the quadratic has no real solutions. If you think about it, if the discriminant is negative, that means the value of the roots obtained from the quadratic equation would be of the form ${-b \pm i\sqrt{x}}/{2a}$, where $x$ is the number remaining when $i$ is factored out of the negative square root.

If the discriminant is zero: the quadratic is one unique real solution. The discriminant would amount to nothing and therefore the single value remaining would simply be ${-b}/{2a}$.

If the discriminant is positive: the quadratic has two unique real solutions. The discriminant would be a positive square root value that would either be added to or subtracted from $-b$, generating two unique values.

When the discriminant $(b^2 - 4ac)$ is positive, square roots of positive numbers are always real, so the solution(s) will be real. However, there can be one solution or two solutions, depending on whether or not the discriminant is equal to 0. In the Quadratic Formula, there's a $+$ or $-$ between the $-b$ and the discriminant, so if the square root of the discriminant is 0, the $+$ or $-$ won't do anything, and there will only be one answer. This also means that if the discriminant is equal to 0, the graph of that quadratic function's vertex is on the x - axis.

Above there are 2 quadratic equations' graphs: a green one and a blue one. Their equations are: $y = x^2/4$ and $y = x^2 - 4$. The functions' solutions are labeled $S$, $X$ and $S$.

Congratulations!

The nature of the roots of a quadratic equation can be determined by taking the discriminant of the equation.

By nature, it is meant as whether the roots are real distinct, real equal or non-real numbers.

The discriminant of the quadratic equation $ax^2 + bx + c = 0$ is $$\Delta = b^2 - 4ac$$ where $\Delta$ refers to the discriminant.

• If $\Delta > 0$, the equation has real distinct roots.
• If $\Delta = 0$, the equation has real equal roots.
• If $\Delta < 0$, the equation has non-real roots, i.e. No real roots.

(Video) Using the Discriminant to Predict the Types of Solutions to a Quadratic Equation

by Patrick JMT

In this video, PatrickJMT states that we will be learning how to predict the types of solutions to a quadratic equation using the discriminant.

$\large{\text{Summary}}$

He states that the discriminant is the term under the square root of the quadratic equation, so $b^2 - 4ac$

If $b^2 - 4ac$ is positive, then you are going to have two real solutions. If its zero, you are going to have exactly one real solution. If its negative, you are going to have two imaginary solutions, which means there will be no real solutions.

He then gives us a quadratic equation to look at: $3x^2 - 2x + 2 = 0$. We want to know the nature of the roots, which means how many real solutions do we have? When breaking the equation down into terms of $a$, $b$ and $c$, we see that the $a$ term is $3$, $b$ is $-2$, and $c$ is $2$.

Since we have these values, we can plug them into the discriminant to get $(-2)^2 - 4(3)(2)$. Performing the arithmetic, we get a discriminant of $-20$. Clearly, this is a negative number, so looking at our rules from before, we see that we are going to have no real solutions to the problem.

This video is a great example of using the discriminant to predict what solutions our quadratic equation will have before we even apply the quadratic formula.

Consider a quadratic $ax^2+bx+c$.

We should knock off the easiest part of this question first. The $y$-intercept occurs when $x=0$. Plugging $x=0$ into your generic quadratic should give you the only $y$-intercept, which is $c$.

But the number of $x$-intercepts is a trickier matter. This happens when $y=0$, and therefore when $ax^2+bx+c=0$. We know need to figure out the number of solutions.

That's right! In general, a quadratic will have two solutions. Of course, you can probably think of quadratics with just one, or no solution. An example would be $x^2=0$ The only solution is $x=0$. But, how? Shouldn't there be two solutions?. It turns out that $x=0$ should really be counted twice, since the equation has a square. The solution $x=0$ has what is called a multiplicity of $2$. Now, let's examine this equation with the quadratic formula, which lets us find the roots of a quadratic. If know this formula, you probably know it boils down to some stuff plus or minus the square root of $b^2-4ac$. You probably know this quantity as the discriminant.

You probably noticed that the discriminant doesn't have to be positive. Let's see what happens when this is the case.

The discriminant of $x^2$ is $0$. But what does this really mean? It means that, in the quadratic formula, the plus or minus part of the formula does not matter. Thus, there is only one solution. In general, if the discriminant is $0$, there is only one solution.

If the discriminant is positive, there are clearly $2$ solutions—the negative and positive.

What if the discriminant is negative? Let's go back to one of our previous quadratics.

Yup. When the discriminant is negative, the square root of the discriminant is not real. Thus, there are no roots in this case. (The roots are actually complex numbers, if you're wondering).

So to summarize:

• Discriminant Positive: $2$ solutions
• Discriminant Zero: $1$ solution
• Discriminant Negative: $0$ solutions

When do Quadratic Equations have real solutions?

Recall that a quadratic equation, written in the form $ax^2 +bx + c$, has the solutions $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

There are two possible ways to make the quadratic formula undefined- setting the denominator to zero, and creating a negative under the radical. The first one of these is trivial- if $a$ equals zero then the expression is no longer quadratic! Thus, the quadratic formula does not apply.

The second is a little more interesting. In that case, the expression does have roots, but they are not real! Instead, the expression has roots that are imaginary. Imaginary roots won't be covered here, but if you are curious, you can learn about imaginary roots here!

Overall, there are three possibilities for the number of roots that a quadratic function may have:

1. Two roots - this is the ordinary case and occurs when $\sqrt{b^2 - 4ac} > 0$ and the $\pm$ applies algebraically. Graphically, this occurs when the quadratic function crosses the $x$-axis twice:

1. One root - this is where $\sqrt{b^2 - 4ac} = 0$ so the $\pm$ has no effect algebraically. Graphically this occurs when the quadratic just barely touches the $x$-axis

1. Zero Roots - this is where $\sqrt{b^2 - 4ac} < 0$ and so the quadratic equation is undefined algebraically. Graphically, this occurs when the quadratic does not touch the $x$-axis at all, thus it is impossible to find such an $x$ such that $y$ is zero.

Hopefully this cleared up when a quadratic has real roots and when it does not. Just remember that if $\sqrt{b^2 - 4ac} \geq 0$ then there will be real roots- otherwise there will not

(Video) Factoring and the Quadratic Formula

by PatrickJMT

Patrick JMT has a great video explaining why there are quadratics with no real solutions and what to do about them.

$\large{\text{Summary}}$

Not all quadratic equations have solutions. At this point in the video, Patrick JMT is giving an example of when a quadratic equation does not have a solution. He gives the example $x^2 + 1 = 0$.

He notes that when you plug in $x=0$ into the equation, you get a final answer of $1 = 0$. Likewise, if you plug any positive or negative number in for $x$, the final answer will never be $0$.

Graphing this equation, the graphed line will never touch or even reach the $x$-axis, which is a strong indication to no solutions for the equation. Rewriting the equation to read $1x^2 + 0x +1$ helps when trying to find the determinant of the equation. $a = 1$, $b = 0$, and $c = 1$. You would find that $b^2 – 4ac$ is equal to $0^2 – 4(1)(1)$, which is $-4$.

This is a big problem, because this determinant is the value that is under the square root in our quadratic equation. Therefore, since the determinant is $<0$, there are no real solutions to this equation.

(Video) The Discriminant and Quadratic Graphs

by HEGARTYMATHS

Here's a video by HEGARTYMATHS showing how the discriminant relates to the number of $x$-intercepts the graph will have.

$\large{\text{Summary}}$

The discriminant is part of the quadratic formula. Its the part inside the radical symbol, $b^2-4ac$. The discriminant can be used to quickly tell how many real solutions a quadratic function will have. This also means how many $x$-intercepts it has. There are three different cases to look at:

1. Case 1: $b^2-4ac>0$ - Here the discriminant is positive. Because of this the quadratic formula will come out with two different answers. So for this case we have two real solutions and the graph will have two $x$ intercepts.
2. Case 2: $b^2-4ac=0$ - Here the discriminant is zero. Because of this the quadratic formula will come out with one answer. So for this case we have one real solution and the graph will have one $x$ intercept; it's vertex.
3. Case 3: $b^2-4ac<0$ - Here the discriminant is negative. Since we can't take the square root of a negative number, we can tell that the quadratic function has no real solutions and no $x$ intercepts. Later, when you learn about complex numbers, you'll learn that there are two imaginary solutions.