Using the Quadratic Formula

Hello, and welcome to this video about using the quadratic formula! Today we’ll learn what the quadratic formula is, when you use it, and how it’s applied to find roots in quadratic equations.

Let’s start by talking about quadratic equations. A quadratic equation in standard form is written as \(ax^{2}+bx+c=0\), where \(a\neq 0\) and \(a\),\(b\), and \(c\) are all real numbers.

What is the Quadratic Formula?

The quadratic formula is an algebraic equation that helps solve any quadratic equation. The formula is:

\(x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\)

By substituting in given values for the variables a,b, and c, you can solve for x using the quadratic formula. Solving for x gives us solutions for the quadratic equation. These solutions are also called roots.

Let’s consider the quadratic equation \(x^{2}+x-6=0\) and use the quadratic formula to find solutions for \(x\).

First, identify the values for the variables a, b, and c in the quadratic equation. Since the first term, \(x^{2}\), is the same as \(1x^{2}\), the value for a is equal to 1. In the second term, x, the value for b is also equal to the coefficient of 1. Since the last term is -6, the value for c is equal to -6.

\(a=1\)
\(b=1\)
\(c=-6\)

Next, substitute the values for a, b, and c into the quadratic formula. So let’s write out our quadratic formula again so that we remember it.

\(x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}\)

And now we’re going to plug in our a, b, and c into this equation.

\(x=\frac{-\left ( 1 \right )_{-}^{+}\sqrt{\left ( 1 \right )^{2}-4\left ( 1 \right )\left ( -6 \right )}}{2\left ( 1 \right )}\)

From here, solve for x using the quadratic formula. So all we have to do is simplify. So we’re going to leave the -1 here for now and move on to the next part. So we have:

\(x=\frac{-\left ( 1 \right )_{-}^{+}\sqrt{ 1-\left ( -24 \right )}}{2\left ( 1 \right )}\)

1-(-24) is the same as saying 1+24, which is 25.

\(x=\frac{- 1 _{-}^{+}\sqrt{25}}{2\left ( 1 \right )}\)
 
\(x=\frac{-1_{-}^{+}5}{2}\)

And now we’re going to break this apart into two different equations. So we’re going to have:

\(x=\frac{-1 +5}{2}\)

Remember, x is equal to this, so this is one of our values for x.

\(x=\frac{4}{2}=2\)

And then for our other value of x, we’re going to do:

\(x=\frac{-1-5}{2}=\frac{-6}{2}=-3\)

 
So the roots for this quadratic equation are 2 and -3.

In the example we just looked at, there were two possible solutions, or roots, for x. Quadratic functions can have two solutions, one solution, or no real solutions for x. To figure out how many solutions a quadratic equation has, look at the discriminant in the quadratic formula. The discriminant is the portion of the quadratic formula that’s beneath the square root symbol. It’s the part that’s read as \(b^{2}-4ac\).

When simplified, the discriminant can be a positive number, a negative number, or zero, and its sign tells us how many solutions the quadratic equation has.

• When the discriminant is positive, then the quadratic equation has two real solutions.
• When the discriminant is negative, then the quadratic equation has no real solutions.
• When the discriminant is zero, then the quadratic equation has exactly one real solution.

In the previous example, the discriminant was 25. Since it’s a positive number, we know that the quadratic equation has two real solutions, which is what we found when we worked it out.

Quadratic Formula Examples

Consider the quadratic equation \(10x^{2}-9x+6=0\). And let’s use the quadratic formula to find solutions for x.

First, we need to identify the values for the variables a, b, and c in the quadratic equation. Our first term, \(10x^{2}\), tells us that a is equal to 10. Our second term, \(-9x\), tells us that the value of b is \(-9\). And our third term, \(+6\), tells us that c is equal to 6.

\(a=10\)
 
\(b=-9\)
 
\(c=6\)

Now we’re going to use the quadratic equation to solve this problem. So our formula is:

\(x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}\)

Now remember, writing out formulas helps you remember it, so whenever you can, it’s a really helpful thing to do. Now all we have to do is plug in our numbers for our variables.

So, -b, b is -9, so be careful when plugging it in. You need to make sure you have the negative -9.

\(x=\frac{-\left ( -9 \right )+\sqrt{\left ( -9 \right )^{2}-4\left ( 10 \right )\left ( 6 \right )}}{2\left ( 10 \right )}\)

Which is equal to, minus a negative is equal to a positive, so we are going to have a positive 9 right here.

\(x=\frac{9_{-}^{+}\sqrt{81-240}}{20}\)

So this gives us:

\(x=\frac{9_{-}^{+}\sqrt{-159}}{20}\)

 
And we can stop right here because we see our discriminant is -159. Negative numbers don’t have real square roots because the square is either positive or 0. So remember, if the discriminant is a negative number, then there are no real solutions to the equation, so we can stop right here. Our answer is that there are no real solutions for x.

Now it’s your turn. I’m going to give you a quadratic equation, and you need to use the quadratic formula to identify possible roots for x

\(9x^{2}+12x+4=0\)

Pause the video here and try it yourself. And when you’re done, resume the video and we’ll go over everything together.

Let’s take a look at this one together. Start by identifying the values for a, b, and c. Based on the equation, we see that \(a=9\), \(b=12\), and \(c=4\).

Next, substitute the values for a, b, and c into the quadratic formula:

So remember, the formula is:

\(x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}\)

And then we just take our numbers up here and plug them in for all our variables. So we have:

\(x=\frac{-\left ( 12 \right )_{-}^{+}\sqrt{\left ( 12 \right )^{2}-4\left ( 9 \right )\left ( 4 \right )}}{2\left ( 9 \right )}\)

So this simplifies to:

\(x=\frac{-12_{-}^{+}\sqrt{\left ( 12 \right )^{2}-4\left ( 9 \right )\left ( 4 \right )}}{18}\)

So then, we have:

\(x=\frac{-12_{-}^{+}\sqrt{0}}{18}\)

\(\sqrt{0}=0\), so all we have is:

\(x=\frac{-12_{-}^{+}0}{18}\)

 
And if you add or subtract 0, that’s not going to change your number at all, so this is really the same as:

\(x=\frac{-12}{18}\)

Which we can then simplify by dividing both the numerator and denominator by 3, so that’ll give us:

\(x=\frac{-12\div 3}{18\div 3}=\frac{-4}{6}\)

And we can simplify that one more time by dividing both by 2 to get:

\(x=-\frac{2}{3}\)

So \(x=-\frac{2}{3}\). And that’s our root for this equation.

Quadratic functions occur in many real-world situations. One example is shooting a cannonball out of a cannon. When you do this, it goes up and then it comes back down in the shape of a parabola. Since a parabola is the graph of a quadratic function, you can use the equation for the path the cannonball follows to answer questions about its time, speed, and distance.

There are lots of other problem-solving instances in which quadratic equations are helpful. Understanding them helps us apply mathematical concepts in the real world.

I have one more problem for you to try. Let’s take a look at a story problem that we can solve by using the quadratic formula:

A high school physics club shoots a model rocket into the air from 12 feet above the ground. The initial velocity of the rocket is 19 feet per second. The equation \(h=-10x^{2}+19x+12\) can be used to show the height of the rocket after x seconds. How long does it take for the rocket to hit the ground?

We can solve this problem by applying the quadratic formula to the equation given in the scenario. This problem is a little bit more challenging, but I know you can handle it! Pause the video here and try it yourself. And when you’re done, we’ll go over it together.

Let’s take a look at this one.

In this scenario, we are looking for the amount of time it takes for the rocket to reach the ground. The height (h) at ground level is 0 feet. Therefore, we can replace 0 for h in the equation. So we can rewrite it as:

\(0=-10x^{2}+19x+12\)

And now it looks like our quadratic equation in standard form, just with our 0 on the left side instead of the right side. So now that we understand that our equation is really just in standard form, we can identify the values for a, b, and c. Based on our equation, we see that \(a=-10\), \(b=19\), and \(c=12\).

Now we can substitute these values into our quadratic formula. Here is the formula:

\(x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}\)

No we can plug in our numbers. So we have:

\(x=\frac{-\left ( 19 \right )^{2}-4\left ( -10 \right )\left ( 12 \right )}{2\left ( -10 \right )}\)

So this is equal to:

\(x=\frac{-19_{-}^{+}\sqrt{361-\left ( -480 \right )}}{-20}\)

So this is equal to:

\(x=\frac{-\left ( 19 \right )+\sqrt{841}}{-20}\)

So this is equal to:

\(x=\frac{-19_{-}^{+}29}{-20}\)

And now from this step, we’re going to take this equation and break it into two separate equations because of this plus or minus. So we’re going to have:

\(x=\frac{-19-29}{-20}\)

And we’re also going to have:

\(x=\frac{-19-29}{-20}\)
 
\(x=\frac{-19+29}{-20}=\frac{10}{-20}=-\frac{1}{2}\)

And over here:

\(x=\frac{-19-29}{-20}=\frac{-48}{-20}\)

Our negatives will cancel each other out, and if we divide both the numerator and denominator by 4, we’ll get:

\(x=\frac{-48}{-20}=\frac{48\div 4}{20\div 4}=\frac{12}{5}\)

So the solutions to this equation are \(x=-\frac{1}{2}\) and \(x=\frac{12}{5}\).

Now that we have the two roots for \(x\), we must interpret what they mean in the context of the word problem. Recall that \(x\) stands for seconds, measuring the amount of time it takes for the model rocket to hit the ground. Time is measured with positive numbers, not negative numbers. Therefore, the negative root of \(x\) does not apply to this scenario. Based on the positive solution for \(x\), the amount of time it will take for the rocket to hit the ground is \(\frac{12}{5}\) seconds.

Great work!

I hope this video about using the quadratic formula was helpful. Thanks for watching, and happy studying!

Quadratic Formula Practice Questions