Admission Test | Only 8% of People Solved This Correctly

[00:00:00]Hello everyone. Happy to see you here.

[00:00:01]Today we have great and quick algebra question. We know that the sum a plus b equal to 14 and a product a times b equal to 20. We need to solve this question for a and b. So right now you can easily pause the video and write your solution in the comments below. It will be really interesting to check your answers in the end of the video. So right now let's start. First of all, we have two equations and I want to look closely at this first equation. Let's express from the first equation, let's express our b. So this is our first step. So from the first equation, b equal to 14 minus a, 14 minus a. And right now let's plug in this b instead of this b, let's plug in 14 minus a to the second equation. So let's do it. So we have a and instead of b we have 14 minus a.

[00:00:48]Okay, so 14 minus a equal to equal to 20. Right now let's simplify it. We can easily multiply our a by this parenthesis. Uh so let's do it. So we have 14a minus a square and equal to 20. And if we look closely like from a different angle, this is a quadratic equation but we prefer a little bit different order. Yeah, we prefer minus a square on the first position. So let's do it. So minus a square on the first position. On the second position we write this plus 14a.

[00:01:22]So plus 14a and minus 20 from the right side. So minus 20 from the right side equal to equal to zero. And the final trick and my quick recommendation because we prefer multiplying both sides by minus one. So we prefer changing the signs to the opposite one because we prefer this a square without this negative sign. Okay, it's much easier for us to simplify.

[00:01:47]Okay, so a square and the next thing minus 14a.

[00:01:51]Uh and plus 20. So this is like a quick trick. I really hope you learn this trick. It's cool. So we just multiply or divide both sides by minus one and then we will basically eliminate this negative sign from here and we have a positive. So we have a classic order of signs.

[00:02:06]Okay, right now we have a basic quadratic equation and of course this is up to you how you prefer solving this type of question. For example, let's write our coefficients quickly. Let's write our a equal to one uh b equal to minus 14 and c equal to 20. So we have our coefficient and we know our formulas. So let's start for example with discriminant uh quickly. We have discriminant equal to b square minus 4 ac. So equal to b square. So minus 14 square minus 4 times 1 and times c equal to >> [snorts] >> equal to 20. So let's simplify it. Minus 14 square this is our table case 196 minus 4 times 20 equal to 80. So our discriminant is equal to 116. And this is really great because discriminant is greater than zero. So from here it implies that we have two two roots. And moreover, we can say that we have two real number roots and we have two pairs of solution. This is our first first thought. Okay, what we're going to do next? Right now let's mm Right now let's write our a first and a second. So let's find our roots because this is only our discriminant, yeah? Uh so let's do it.

[00:03:25]So one second, yeah, like that, like that. Yeah.

[00:03:28]Uh so we have minus b plus minus square root of discriminant and all over all over 2a. Okay, all over 2a. So right now let's plug in everything. We know everything from here. So minus b minus minus 14. So minus minus 14 plus minus square root of discriminant square root of 116.

[00:03:52]>> [snorts] >> And all over we're going to divide all of this by two times two times one. So equal to Right now minus minus 14 equal to 14 plus minus square root of 116. Let's write it like that without any changes.

[00:04:09]So square root of 116 and all over two.

[00:04:13]And right now a few thoughts about this 116. So this is not a table case but let's do really interesting and quick trick. So 116 right here we divide by two. So if it's not a table case, my quick recommendation is to divide it step by step by two or by three. So let's divide it by two.

[00:04:32]And we have 58.

[00:04:34]Let's divide it by two. We have 29. So basically we can express this 116 like that. So we have 14 plus minus square root of we're going to express this as two times two we have four. So let's write it like four times 29 all over all over two. And right now square root of a product equal to a product of square roots. So we [snorts] can split it. We can write it as plus minus square root of four times square root of 29. And right now from this angle square root of four equal to equal to two. So as a result we have 14 plus minus two square root of 29 over over two. And our last step Of course there are two ways how can we do that.

[00:05:23]We can factor two from our numerator or we can write this one as 14 over two and this one over two. So we can basically we can cancel everything by two. So we have seven plus minus two square root of uh plus minus square root of 29 without two. So square root of 29.

[00:05:41]So this is our answer. But one really interesting thing right here we have two we have two roots. So this is our a first. Let's write with the positive sign. So seven plus square root of 29.

[00:05:53]And we also have negative one. So a second equal to seven minus square root of 29. But this is not our answer. This is not our pair because in the beginning we had our b as well. So b equal to 14 minus a. So right now let's solve it for right here let's solve it for b first.

[00:06:15]So b first equal to 14 minus a first. So 14 minus a first.

[00:06:22]And right here b second equal to 14 minus a second, yeah? Like that. And right now let's plug in everything. So b first equal to 14 minus a first. So seven plus square root of 29. From here we have b first equal to 14 minus seven and minus square root of 29. So basically we have absolutely the same answer in terms of coefficients. So we have seven minus square root of 29. We have a symmetrical pair. We have seven plus square root of 29 and seven minus square root of 29. And using the same logic we can solve it for b second. But basically we have absolutely the same thing as right here. So b second will be equal to seven plus square root of 29.

[00:07:09]But maybe a lot of students don't trust me. So let's do it like that. So 14 minus a second equal to seven minus square root of 29. So basically b second equal to 14 minus seven and plus square root of 29. So from here 14 minus seven equal to seven. So b second equal to seven plus square root of 29. So basically we have two pairs. So I want to write it as the full answer and then we will check it.

[00:07:38]So our answer to this question So we have first pair. Not like a first a second. But our pair looks like that.

[00:07:45]A first a second. So we have seven plus square root of 29 and seven minus square root of 29. And another parenthesis seven minus square root of 29. And seven plus square root of square root of 29. So right now let's check it. So we have our answers.

[00:08:07]So right now let's let's check it. Let's let's prove our answer. First of all, I'm going to write our question from the beginning. So a plus b equal to 14 and ab equal to equal to 20. And right now one really interesting trick because we have addition and multiplication. So basically we don't need to check both of these parenthesis. We we don't need We need to check only for example this one or this one because a plus b equal to b plus a. It doesn't matter what we're going to add to this or this expression to this one or this to this. We have absolutely the same thing because we have a symmetrical pairs, symmetrical answers and the same coefficients. So first of all let's check real quick our addition. So so a plus b equal to So seven plus square root of 29 plus seven and minus square root of 29. Right here we're going to cancel this square root sign. Seven plus seven equal to equal to 14. So basically we have the thing right here. So a plus b equal to 14. Absolutely correct correct expression. And in the end let's check real quick our multiplication. So we have a times b equal to So seven plus square root of 29 and seven minus square root of 29. And one really interesting moment. Of course you can multiply parenthesis by parenthesis. But right here we have difference of squares. I hope you you remember this formula from school. When we have like x plus y, x minus y. On the left side we can write it as difference of squares x square minus y square. This this is a classic school identity. I hope you understand it. So this is our difference of squares. Seven square minus square root of 29 to the power two. And from here >> [snorts] >> this is equal to 7 squared 49 minus there here we can cancel it. 49 minus 29 equal to 20. So expression that we exactly need on the left side and this is our answer. This is our answer to this question. Our answer is absolutely correct because we check it, we prove it right here on the bottom. So I really hope you understand it. But if you still have any question, write your question in the comments below. It's a classic um easy question from from from the beginner level. Sometimes I see this type of question on entrance examination, um you know, and this classic question for beginners. And of course, if you studying a math, if you have a problems with math, this is really great question to practice quadratic equation, to practice algebraic skills, and to practice different identities, very classic and basic uh question. So thank you everyone for your time. Take care of yourself.

[00:10:51]Have a great day. Write your thoughts about it. Write your notes. What do you think about it? Write your respond in the comments below. It will be really interesting to read about it. So thank you everyone for your time. Take care of yourself and have a great day. See you in the next videos.