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. Read full transcript
There's another method called completing the square which I've 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 the 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 formula, and 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 plus bx plus c equals zero.
So notice that is a quadratic equation set equal to zero, 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 plus-minus 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 that you will, this 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 minus 4ac. This is sometimes call the discriminant. That's a term that you do not need to know for the ACT. But that expression, b squared minus 4 ac, is very important.
The reason is, if it's positive, then we get two real square roots, and then we're going to wind up with two different values for x. It's going to be negative b plus something, and negative b minus something over 2A. We're going to get two roots. In rare cases when b squared minus 4ac equals 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 minus 4ac equals zero. 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 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 going to 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 Formula. It is quadratic equation. It's already in standard form. It's already set equal to zero.
And we're gonna solve for x. It's not immediately obvious how we would factor that or use completing the square, so the quadratic formula is actually not a bad choice with this equation. So we can see that a equals one, b and c equal negative one, 1. Very important to remember those negative signs when you're reading off a, b and c of 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 five. And so we have two roots here. The two roots, one of these is one plus root five over two. The other is one minus root five over two. 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 going to solve this, not 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 four. 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 solve it that way, but let's solve it with the quadratic formula.
So we get a equals one, b equals negative 12, c equals 31. Plug in all those numbers, and four times 31, well, four times 30 is 120, so four times 30, rounded to 124. We subtract that, we get 12 plus root 20 over two. Now, remember the lessons that we had on radicals, we can simplify 20 because 20 is four times five.
And four is a perfect square, so we can separate that into square root of four times square root of five, and the square root of four is two. Well now everything in the numerator is divisible by tow, so we can cancel the two, and we get six plus or minus root five. And that is the solution to this equation, six plus root five and six minus root five, 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 zero, so we just have to subtract three from both sides and get it equal to zero. All right. 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, 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 an other method of solution is easier for you. But here I'll demonstrate the quadratic formula. We plug everything in. Of course, four times seven is 28, 36 minus 28 is eight.
Square root of eight can be simplified because eight is four times two so square root of eight is square root of four times square root of two or in other words two root two. Then we can divide everything by two and we get three plus or minus root two. 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 emphasize once again, it's not your only option, and 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 in the quadratic formula.