90% Get This Math Olympiad Wrong! | Nice Square Root Problem

Added: 2026-05-05

[00:00:01]Hello, you're welcome. Let's just solve this nice problem √30 + √10 all over √30 - √10.

[00:00:10]Solution here.

[00:00:11]Let's use two two methods here.

[00:00:15]The first method.

[00:00:18]Now, what is given here? We can write 30 and we have √ 10 * 3 then plus √10.

[00:00:29]Also divided by here we have √ 10 * 3 - √10.

[00:00:39]Then from here, this follows. When we have √ a * b, we can separate this as √ a * √ b.

[00:00:47]Then also from here we have √10 * √3 + √10 divided by Here also we have √10 * √3 - √10.

[00:01:09]Then from here, √10 is common factor it out.

[00:01:13]We have √10 into bracket, here remains √3 plus one into bracket also divided by √10 is common here. We have √10 into bracket, we have √3 - 1 left.

[00:01:33]That is Here, √10 cancel each other.

[00:01:36]We're left with √3 plus one divided by √3 - 1.

[00:01:46]Then from here, I can rationalize this sort and multiply by the conjugate of the denominator, which is √3 + 1.

[00:01:53]So, multiply numerator and denominator by this conjugate sort. We have √3 plus one multiplied by √3 plus one and divided by also here we have √3 - 1 multiplied by √3 - 1.

[00:02:20]The next step same same multiplying when we have a * a, write it as a squared.

[00:02:26]This a, I write as √3 plus one all squared which is divided by >> [snorts] >> Here, this follows. This is plus, excuse me. Well, when we have a - b * a + b I write this as a squared - b squared.

[00:02:48]This also from here I write this as √3 squared - 1 squared.

[00:02:56]Then we expand this bracket and this follows when we have a + b all squared, which can be written as a squared plus b squared plus 2ab.

[00:03:10]That is here we have √3 squared plus 1 squared plus 2 * √3 * 1 all over Here, square cancels square root. We have 3 minus 1 squared as 1.

[00:03:26]That is also square cancels square root here. This gives us √3 plus one. 1 squared as 1 then plus 2 * √3 * 1 that's still 2 √3 and over 3 - 1 that's 2.

[00:03:42]>> [snorts] >> That is we have 3 + 1 4 and plus 2 √3 all over 2.

[00:03:51]The next step 2 is common up here. We have 2 into bracket 2 left here and plus √3 and over 2.

[00:04:03]2 cancel each other here and this becomes 2 plus √ 3.

[00:04:09]So, here we have this as the simplified form of the given problem. Then let's use the second method.

[00:04:17]From what is given here, we can rationalize this directly. Multiply the numerator and denominator by the conjugate sort of the denominator that is √30 plus √10.

[00:04:27]So, use this to multiply both sides. So, we have √30 plus √10 multiplied by also √30 plus √10.

[00:04:44]Then divided by Here also we have √ 30 - √10 multiplied by √30 plus √10.

[00:04:59]Then here also same same multiplying when we have a * a which is same thing as a squared.

[00:05:05]Can express this as well as √30 plus √10 all squared and divided by here follows when we have a - b into bracket, open bracket a + b which is same thing as a squared - b squared.

[00:05:26]That is here we have √30 squared - √10 squared.

[00:05:34]Then [snorts] here we expand this. Also when we have a + b all squared which is same thing as a squared plus b squared plus 2ab.

[00:05:46]Then >> [snorts] >> here we have √30 squared plus √10 squared plus 2 * √30 * √10 all over Here, square cancels square root. That's 30 minus square cancels square root here. That's 10.

[00:06:08]Then here also square cancels square root. We have 30 plus also square cancels square root 10 and plus 2 * We can bring it together when we have √ a * √ b which is same thing as √ a * b.

[00:06:24]So, here we have √ 30 * 10 and over 30 - 10 which is 20.

[00:06:34]This here also 30 + 10 40 plus 2 * square root of I can write this as 3 * 10 and * 10 here over 20.

[00:06:46]And also here when we have √ a * a which is same thing as a.

[00:06:51]√10 * 10 that's 10. We take 10 out and this becomes 40 plus 2 * 10 then * √3 left over 20.

[00:07:03]Which implies we have 40 plus * 10 that's 20 √3 over 20.

[00:07:11]Then 20 is common. I can factor it out and we have 20 into bracket This remain two then plus √3 close bracket over 20.

[00:07:26]This 20 cancel each other now and this becomes 2 plus √3.

[00:07:32]Which is same thing with what we have in the first method.

[00:07:35]And therefore this given problem here >> [snorts] >> conclude that it simplify as 2 plus √3.

[00:07:49]Don't forget these steps.

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[00:07:57]And see you in the next class.

[00:07:59]Bye for now.

[00:08:05]Hello, you're welcome. Let's solve this nice exponential equation to find the value of x here. Solution We take 3 raised to power 6 to the left hand side and we have x over 2 all raised to power 6 - 3 raised to power 6 equals to zero here.

[00:08:26]Then we can write this as x over 2 all raised to power 3 then all raised to power 2 as 2 * 3 equals 6 here minus also here we have 3 raised to power 3 then all raised to power 2 equals to zero here.

[00:08:45]Next step, this follows when we have a squared - b squared This is same thing as a + b into bracket, also into bracket a - b.

[00:08:59]That is here a is standing as x over 2 all raised to power 3 and b is standing as 3 raised to power 3.

[00:09:10]Then we can write this equation as x over 2 all raised to power 3 plus 3 raised to power 3 into bracket also into bracket x over 2 all raised to power 3 - 3 raised to power 3 close bracket equals to zero here.

[00:09:39]>> [snorts] >> Then we have two possible cases here. The first one x over 2 all raised to power 3 plus 3 raised to power 3 equals to zero here, excuse me.

[00:09:55]Or we have x over 2 all raised to power 3 minus 3 raised to power 3 equals to 0 here.

[00:10:07]Then solving from the first case, this follows when we have a raised to power 3 plus b raised to power 3 is same thing as a plus b into bracket open bracket a squared minus ab plus b squared.

[00:10:27]Then we can rewrite this as x over 2 that's a plus b is 3 then into bracket a squared that's x over 2 all squared minus 3 times x over 2 plus 3 squared close bracket equals to 0 here.

[00:10:53]Then yeah this becomes x over 2 plus 3 into bracket open bracket expansion here give us x squared over 4 minus 3 x over 2 then plus 3 squared which is 9 close bracket equals to 0 here.

[00:11:21]Then yeah also we have two possible cases x over 2 plus 3 equals to 0 or we have x squared over 4 minus 3 x over 2 plus 9 equals to 0 here.

[00:11:40]That is solving on this side and take 3 to right hand side we have x over 2 equals to minus 3.

[00:11:51]Then multiply both side by 2 that is 2 times x over 2 equals to 2 times minus 3. 2 cancel each other here we have x equals to 2 times minus 3 that's minus 6 so we have x equals to minus 6 here which is a real solution.

[00:12:10]Also solving here we can remove this fraction multiply through by the LCM which is 4.

[00:12:19]Then here we have x squared now minus 3 times 6 x plus 36 equals to 0 here.

[00:12:30]Yeah we quadratic equation now where a equals to 1 b equals to minus 6 and c equals to 36.

[00:12:41]Then apply the quadratic formula which is x equals to minus b plus or minus square root of b squared minus 4 ac all over 2 a.

[00:12:57]All we have here becomes x equals to minus minus 6 plus or minus square root of minus 6 squared minus 4 times 1 times 36 all over 2 times 1.

[00:13:18]Then here we have x equals to minus times minus turns to plus that's 6 plus or minus square root of minus 6 squared that's 36 minus 4 times 1 times 36 we can see that this is 4 times 36 then all over 2.

[00:13:37]>> [snorts] >> And 36 is common here that is x equals to 6 plus or minus square root of factor 36 out as 36 into bracket we have 1 minus 4 left here all over 2.

[00:13:53]Then [snorts] this becomes x equals to 6 plus or minus square root of 36 times 1 minus 4 that's minus 3 all over 2.

[00:14:05]Then we separate this and we have x equals to 6 plus or minus root 36 that's 6 times minus 3 that's same thing as root 3 i over 2.

[00:14:19]Then 2 is common here factor it out we have x equals to 2 into bracket 3 plus or minus 3 root 3 i then all over 2.

[00:14:34]Then here 2 cancel each other we have x equals to 3 plus or minus 3 root 3 i. Also we have two complex solutions here.

[00:14:48]Then solving from the second case here this also follows when we have a raised to power 3 minus b raised to power 3 which is same thing as a minus b into bracket open bracket a squared plus ab plus b squared.

[00:15:06]Then this also becomes x minus 2 minus 3 into bracket open bracket x over 2 all squared plus 3 times x over 2 plus 3 squared close bracket equals to 0 here.

[00:15:31]Then we have x over 2 minus 3 into bracket open bracket expansion give us x squared over 4 plus 3 x over 2 plus 9 close bracket equals to 0 here.

[00:15:50]That is we also have two possible cases x over 2 minus 3 equals to 0 or we have x squared over 4 plus 3 x over 2 plus 9 equals to 0 here.

[00:16:07]Then take minus 3 to this side it becomes plus that's x over 2 equals to 3.

[00:16:14]Then this 3 over 1 cross multiply x times 1 that's x equals to 2 times 3 which is 6 which is a real solution here.

[00:16:24]Then here also we clear this fraction multiply through by 4 here we have x squared plus 6 x plus 36 equals to 0 here.

[00:16:37]Then we also have a quadratic equation where a equals to 1 b equals to 6 and c equals to 36.

[00:16:48]Apply the quadratic formula.

[00:16:51]Here also we have x equals to minus 6 plus or minus square root of 6 squared minus 4 times 1 times 36 all over 2 times 1 which implies we have x equals to minus 6 plus or minus square root of 6 squared that's 36 minus 4 times 1 times 36 is same thing as 4 times 36 all over 2.

[00:17:23]Then next step 36 is common here factor it out we have x equals to minus 6 plus or minus square root of 36 into bracket 1 minus 4 left here all over 2.

[00:17:39]That is we have x equals to minus 6 plus or minus square root of 36 times minus 3 all over 2.

[00:17:51]Now we separate this take root 36 as 6 we have x equals to minus 6 plus or minus 6 root minus 3 that's root 3 i all over 2.

[00:18:04]Then 2 is common here factor it out we have x equals to 2 into bracket we have minus 3 plus or minus 3 root 3 i close bracket over 2 a.

[00:18:18]Then 2 cancel each other it becomes x equals to minus 3 plus or minus 3 root 3 i. Also we have two complex solutions here and therefore altogether in this problem we have six solutions here two real solutions and four complex solutions two here and two here.

[00:18:44]And thank you for watching. Don't forget to subscribe for more videos. Turn on the notification bell and share this video with thumbs up and put your comments.

[00:18:52]See you next class and bye for now.