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## Quadratic Formula

Now we'll talk about the Quadratic Formula. Now, of course, this is out of place in some sense because, mostly when we're talking about quadratics, that was back in the algebra section. We were talking about factoring. Usually, if you have to solve a quadratic, the best method will be factoring it, but, sometimes, quadratic equations cannot be factored, and at least one option for these is to use the Quadratic Formula.

There's another method called completing the square, which I have demonstrated in a few lessons. If the quadratic resembles the square of a binomial, and you should know those patterns, the square of the sum, the square of a difference, if it's close to one of those patterns, it's often pretty easy to solve it without the Quadratic Formula.

However, there are some quadratics that are not factorable, and you can't really fit it very easily into one of those neat algebraic patterns, so then the Quadratic Formula is often your best bet. So this is the Quadratic Formula. First of all, keep in mind, it is an If, Then statement. If ax squared + bx + c = 0, so notice that is a quadratic equation set equal to 0, so that's in standard form.

If we put that quadratic into standard form and read off the coefficients, then the solution for x will follow that familiar pattern, that familiar formula. Notice, also, there's a +/- sign in that formula, typically, we'll get two roots. Remember that the graph of a quadratic is a parabola, and many parabolas intersect the x-axis twice, so that's why you'd get two roots. You do not, do not need to memorize this formula.

The test will always provide this formula if you need it, and, once again, most of the time, with a quadratic, your best bet is to factor it. You don't need this formula. The only time the test will give you this formula if something is utterly unfactorable, so just keep that in mind. In particular, notice the expression under the radical in the quadratic formula, b squared- 4ac.

This is sometimes called the discriminant. That's a term you do not need to know for the ATC, but that little expression, b squared- 4ac, that's very important. The reason is, if it's positive, then we get two real square roots, and then we're gonna wind up with two different values for x. It's gonna be negative b plus something and negative b minus something over 2a.

We're gonna get two roots. In rare cases, when b squared- 4ac= 0, then the quadratic has one real root. And you'll notice, for any perfect square, if you go back and look at the square of a sum or square of a difference from the algebra section, all of these obey this condition that b squared- 4ac = 0. So this would be a parabola that is simply tangent to the x-axis at its vertex, so it has only one solution, and, of course, this expression can also be negative.

Well, think about that. If it's negative, this is something under the square root, so we have a square root of a negative. That would be an imaginary number. We'd get two imaginary solutions, and, of course, as always, what you'd get is some real number plus or minus some imaginary number.

The two roots would be two complex conjugates. It is very unlikely that the ACT is gonna give you a quadratic formula that's going to wind up having an imaginary root, but it could happen. So it's just something to keep in mind. Here's a very simple example. So there's a quadratic equation.

It's already in standard form. It's already set equal to zero, and we're gonna solve for x. So it's not immediately obvious how we would factor that or use completing the square, so the quadratic formula's actually not a bad choice with this equation. So we can see that a = 1, b and c = -1.

Very important to remember those negative signs when you're reading off a, b and c for the quadratic formula. So the quadratic formula, which the test would give us, we'd plug in these values. Under the radical, we get a root 5, and so we have two roots here. The two roots, one of these 1 + root 5 over 2. The other is 1- root 5 over 2.

Incidentally, that first root is the golden ratio, and the second root is the reciprocal of the golden ratio, but you do not need to know that for the ACT. Here's another example. We're gonna solve this. Now, notice this one is not in standard form, so step one is always put things in standard form, so we have to put it in standard form.

We subtract 12 from both sides, we subtract 4. As it turns out, if I were gonna solve this, I would actually say that this is very, very close to a completing the square problem, and I would it solve it that way, but let's solve it with the quadratic formula. So, we get a = 1, b = -12, c = 31. Plug in all those numbers, and 4 times 31, well, 4 times 30 is 120, so 4 times 31 is 124.

We subtract that. We get 12 + root 20 over 2. Now remember the lessons that we had on radicals. We can simplify 20 because 20 is 4 times 5, and 4 is a perfect square, so we can separate that into square root of 4 times square root of 5, and the square root of 4 is 2.

Well, now everything in the numerator's divisible by 2, so we can cancel the 2, and we get 6 plus or minus root 5, and that is the solution to this equation. 6 + root 5 and 6 + root 5, those are the two roots. Here's a practice problem, pause the video, and then we'll talk about this. Okay, so this is in the form that the ACT would state it, they give us the quadratic formula, they state all that very clearly, and then they ask us to solve the problem.

Now, of course, they did slip us the trick here. They gave us an equation that's not in standard form, it's not equal to 0, so we just have to subtract 3 from both sides and get it equal to 0. So now we have something in standard form. Incidentally, if you wanted to solve this with completing the square, if you wanted to add whatever you needed to add to get the perfect square, the square of a difference.

That's also a perfectly valid way to solve it. I'll just point out, don't feel compelled to use the quadratic formula if another method of solution is easier for you, but here I'll demonstrate the quadratic formula. We plug everything in. Of course, 4(7) is 28, 36- 28 is 8.

The square root of 8 can be simplified because 8 is 4 x 2, so the square root of 8 is the square root of 4 times square root of 2, or in other words 2 root 2. Then we can divide everything by 2, and we get 3 +/- root 2. So that's the solution to this particular equation. We go back to the problem, and we select answer choice B. In summary, we can find the solution of an unfactorable quadratic using the quadratic formula.

Now I wanna emphasis once again, it's not your only option. In fact, completing the square is often a much quicker, more efficient option, but you certainly can use the quadratic formula. When you need to use the quadratic formula, the ACT will always supply it. You do not need to memorize it. You must make sure that the quadratic equation you are solving is in standard form before you read a, b,and c to plug into the quadratic formula.

Read full transcriptThere's another method called completing the square, which I have demonstrated in a few lessons. If the quadratic resembles the square of a binomial, and you should know those patterns, the square of the sum, the square of a difference, if it's close to one of those patterns, it's often pretty easy to solve it without the Quadratic Formula.

However, there are some quadratics that are not factorable, and you can't really fit it very easily into one of those neat algebraic patterns, so then the Quadratic Formula is often your best bet. So this is the Quadratic Formula. First of all, keep in mind, it is an If, Then statement. If ax squared + bx + c = 0, so notice that is a quadratic equation set equal to 0, so that's in standard form.

If we put that quadratic into standard form and read off the coefficients, then the solution for x will follow that familiar pattern, that familiar formula. Notice, also, there's a +/- sign in that formula, typically, we'll get two roots. Remember that the graph of a quadratic is a parabola, and many parabolas intersect the x-axis twice, so that's why you'd get two roots. You do not, do not need to memorize this formula.

The test will always provide this formula if you need it, and, once again, most of the time, with a quadratic, your best bet is to factor it. You don't need this formula. The only time the test will give you this formula if something is utterly unfactorable, so just keep that in mind. In particular, notice the expression under the radical in the quadratic formula, b squared- 4ac.

This is sometimes called the discriminant. That's a term you do not need to know for the ATC, but that little expression, b squared- 4ac, that's very important. The reason is, if it's positive, then we get two real square roots, and then we're gonna wind up with two different values for x. It's gonna be negative b plus something and negative b minus something over 2a.

We're gonna get two roots. In rare cases, when b squared- 4ac= 0, then the quadratic has one real root. And you'll notice, for any perfect square, if you go back and look at the square of a sum or square of a difference from the algebra section, all of these obey this condition that b squared- 4ac = 0. So this would be a parabola that is simply tangent to the x-axis at its vertex, so it has only one solution, and, of course, this expression can also be negative.

Well, think about that. If it's negative, this is something under the square root, so we have a square root of a negative. That would be an imaginary number. We'd get two imaginary solutions, and, of course, as always, what you'd get is some real number plus or minus some imaginary number.

The two roots would be two complex conjugates. It is very unlikely that the ACT is gonna give you a quadratic formula that's going to wind up having an imaginary root, but it could happen. So it's just something to keep in mind. Here's a very simple example. So there's a quadratic equation.

It's already in standard form. It's already set equal to zero, and we're gonna solve for x. So it's not immediately obvious how we would factor that or use completing the square, so the quadratic formula's actually not a bad choice with this equation. So we can see that a = 1, b and c = -1.

Very important to remember those negative signs when you're reading off a, b and c for the quadratic formula. So the quadratic formula, which the test would give us, we'd plug in these values. Under the radical, we get a root 5, and so we have two roots here. The two roots, one of these 1 + root 5 over 2. The other is 1- root 5 over 2.

Incidentally, that first root is the golden ratio, and the second root is the reciprocal of the golden ratio, but you do not need to know that for the ACT. Here's another example. We're gonna solve this. Now, notice this one is not in standard form, so step one is always put things in standard form, so we have to put it in standard form.

We subtract 12 from both sides, we subtract 4. As it turns out, if I were gonna solve this, I would actually say that this is very, very close to a completing the square problem, and I would it solve it that way, but let's solve it with the quadratic formula. So, we get a = 1, b = -12, c = 31. Plug in all those numbers, and 4 times 31, well, 4 times 30 is 120, so 4 times 31 is 124.

We subtract that. We get 12 + root 20 over 2. Now remember the lessons that we had on radicals. We can simplify 20 because 20 is 4 times 5, and 4 is a perfect square, so we can separate that into square root of 4 times square root of 5, and the square root of 4 is 2.

Well, now everything in the numerator's divisible by 2, so we can cancel the 2, and we get 6 plus or minus root 5, and that is the solution to this equation. 6 + root 5 and 6 + root 5, those are the two roots. Here's a practice problem, pause the video, and then we'll talk about this. Okay, so this is in the form that the ACT would state it, they give us the quadratic formula, they state all that very clearly, and then they ask us to solve the problem.

Now, of course, they did slip us the trick here. They gave us an equation that's not in standard form, it's not equal to 0, so we just have to subtract 3 from both sides and get it equal to 0. So now we have something in standard form. Incidentally, if you wanted to solve this with completing the square, if you wanted to add whatever you needed to add to get the perfect square, the square of a difference.

That's also a perfectly valid way to solve it. I'll just point out, don't feel compelled to use the quadratic formula if another method of solution is easier for you, but here I'll demonstrate the quadratic formula. We plug everything in. Of course, 4(7) is 28, 36- 28 is 8.

The square root of 8 can be simplified because 8 is 4 x 2, so the square root of 8 is the square root of 4 times square root of 2, or in other words 2 root 2. Then we can divide everything by 2, and we get 3 +/- root 2. So that's the solution to this particular equation. We go back to the problem, and we select answer choice B. In summary, we can find the solution of an unfactorable quadratic using the quadratic formula.

Now I wanna emphasis once again, it's not your only option. In fact, completing the square is often a much quicker, more efficient option, but you certainly can use the quadratic formula. When you need to use the quadratic formula, the ACT will always supply it. You do not need to memorize it. You must make sure that the quadratic equation you are solving is in standard form before you read a, b,and c to plug into the quadratic formula.