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Cambridge Interview Question | Test Your Algebra Skills

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mathsmood

mathsmood

7 ай бұрын

Be careful!
Hello. In this video, i am going to show you a great Algebra question that you can use to practice your math skills. On this channel, you will get best math problems, questions and solutions to improve your math skills. Math for fun.
Subscribe to my KZbin channel. Thank you and have a good day:))

Пікірлер: 6
@tacemus
@tacemus 7 ай бұрын
Fun and interesting. Thank you.
@Qermaq
@Qermaq 7 ай бұрын
What a fun one. The trick with these is knowing this must be factorable to something manageable, or else they wouldn't ask it. But how manageable is always the question. This one made me go down a blind alley, but from that alley I could see where I really needed to go, so it was worth the trip. If 53 - 10(sqrt6) is a perfect square, its square root will probably be in the form a + b(sqrt6). Let's assume that for now. The coefficient of the square root will be 2ab. So all the (a,b) pairs are the factors of -10/2 = -5. So we can have 1 and -5 in either order, or -1 and 5 in either order. ab = -5, and a^2 + 6b^2 = 53. Guess and check, we find these don't work. So our assumption that root6 is the radical in the square root was wrong. So next we consider root6's factors which are root2 and root3. Let's say root2 is part of the radical coefficient, and root3 is our actual root. so now ab = -5root2. (5root2)(-1) works, (5root2)^2 = 50, and 3 times (-1)^2 is 3, and lo and behold, 5root2 times -1 works, and we get the solution (5root2 - root3). And we could have made the opposite assumption about which root factor was the root and part of the coefficient and gotten here as well. I don't think I want to attend Cambridge.
@Hanible
@Hanible 7 ай бұрын
5sqrt(2)-sqrt(3)
@user-jq4gy2sh3z
@user-jq4gy2sh3z 7 ай бұрын
Thank you for this video!
@mathsmood
@mathsmood 7 ай бұрын
Thank you for watching and supporting my channel! 🙏 Have a great day. Much love and respect!️❤️
@vladimirkalenyuk5084
@vladimirkalenyuk5084 7 ай бұрын
Maybe a dumb question, but what's the point of such a "simplification"? You still have 4 exactly the same operations to compute the result as in the initial expression: 1 multiplication, 1 negotiation and 2 square roots.
tricky algebra problem, be careful
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