SM2 13.6-3: Solving a Non Factorable Quadratic on a Secant Tangent

Added: 2026-05-09

[00:00:00]Hello and welcome. So we are given a circle with  one secant and one tangent line intersecting it. Now in case you didn't know what those  were, a secant line is essentially a line that intersects with an object twice, right?  Our object being a circle in this case. And a tangent line is the exact same thing but instead  of intersecting twice it only intersects once.

[00:00:22]Okay. So once we know that then it's just a matter  of using your relationship which in this case will be part times the whole equals part times the  whole. Okay, and this is referring to one line each. Now part specifically is referring to the  part that's outside, oops, need to spell that correctly, outside of the circle. All right, let's look at our uh blue  secant line first and make the tangent a little bit easier to think about. So when we're looking  at this one we're looking at part, the outside part of the line specifically. It starts at L  it'll end at D. So the first part of it outside the circle is from L to C and that'll be 7.  All right. Now the whole doesn't refer to the remaining portion of the line. It refers to the  whole shebang from L to D. So that's not just 13.

[00:01:16]That's 7 plus 13. That is the whole segment from  the initial point L, right, to the last point D.

[00:01:25]Now the tangent line is a little bit different,  not by much. So it still starts at L and then to that first intersection point. That whole  section is c plus 8. Do you need parentheses? Yes, make sure you keep them right. And then from  there, well where's the second intersection point, right? There's no other point over there that's  mystically on the circle, right? The whole segment from the initial to the end on the intersection  is still c plus 8. So if you have a tangent line that's essentially multiplying the same thing  twice. That will always be the case. All right. Now you'd say all right well now you just simplify,  right? 13 plus 7 that is 20 right there and then 7 times 20 is 140. And then on the other side we  have c plus 8 times c plus 8. So c times c is c squared, plus 8c, plus 8c, plus 64. All right. And  now we have 140 equals c squared plus 16c, combine your like terms, plus 64.

[00:02:33]I'm going to subtract that 140. We want to make  sure we get everything on the same side because as soon as you see that square that makes this  a quadratic. Solve using quadratic methods, right? For the sake of time this video I'm going  to actually do a little bit of a spoiler for you.

[00:02:47]There is no factors for this one. That's gonna,  it's not gonna factor nicely. Which means that when we have this we're going to have to use one  of our other solving methods other than factoring.

[00:02:58]So that's going to be either completing the  square or solving by the quadratic formula.

[00:03:03]I personally like solving by quadratic  formula. So what we're going to do is we got our c equals negative b plus or minus  square root b squared minus 4ac all over 2a.

[00:03:18]All right. Now from there remember that your  a is essentially the number that's in front of your c squared which is just 1 in this  case. So you have 1 plugged in two places, and then you have your 16 which is your b  in two places, and then you also have your c which is negative 76 in one place. So we  plug that in. So we have c equals negative 16 plus or minus the square root of c (Mistake, meant 16) squared minus 4 times your a times your a times your c, all over 2a.

[00:03:57]Okay. So we have negative 16 plus or minus.  Now this is just calculator plug and chug. All that stuff inside of the square root is going to  end up giving us 560, and that'll all be all over 2 right there. Okay. Now keep in mind that  our directions tell us that we're going to be rounding to a decimal. Does that mean we have  to simplify this? No. I mean you can if you want to but we know it's not going to be factorable  because I told you wink wink. Okay. And then from there we know that because 560 is not a perfect  square that means it's not factorable. Okay. And we know we don't have to simplify it. Does it  simplify? Yes. Do we need to? No. All right.

[00:04:40]So turn to your handy dandy calculator. Plug  it in. Now remember you have to do it twice, one for the plus, one for the minus. When  we do you notice we get two answers. We have negative 19.832 and we have 3.832. So  those are your two solutions. Now are you done?

[00:04:59]Almost. You have not quite finished yet. I think  that was eight three two I said uh 19.832. Okay.

[00:05:10]Once you have those two solutions now it's a  matter of making sure that your solutions work.

[00:05:14]Remember your original goal is to solve for c  and c is a part of that tangent line section.

[00:05:20]Specifically it's a part of a length. Lengths  cannot be negative, right? That doesn't mean that c can't be negative. It just means that lengths  cannot be negative, right? So we plug [whoosh sound] our first answer in 3.832. Well if you have 3.832 plus 8  that's 11.832. That's a positive number, right?

[00:05:43]That means it's good to go. Done. Okay. Next is  to try your negative 19. So if you have negative 19.832 plus 8 that's going to give you a negative  11. Now negative is a problem. You can't have negative lengths which means that your solution is  a bust, right? You can't have that. You're done.

[00:06:04]And that means that 3.832 is your final answer  for this one. All right. Thanks for watching.