Radical Equation Looks Complicated… Until You See This! 😱

追加: 2026-05-07

[00:00:02]In this video, we'll be solving one very, very interesting question from radical equations, and our first move is to take our RHS to the LHS.

[00:00:13]And we can write equation 1 - 1 over 2 - x plus square root of x ^ 2 - 4x + 6 over square root of x ^ 2 - 2x + 3 equal to 0.

[00:00:39]Now, we are going to take -1 common from the denominator 2 - x.

[00:00:45]And minus times minus becomes plus, so we can write 1 plus 1 over x - 2 plus square root of x ^ 2 - 4x We'll split 6 as 4 + 2.

[00:01:10]over square root of x ^ 2 - 2x We'll split 3 here and write 1 + 2.

[00:01:21]will be equal to 0.

[00:01:25]Now, we have to take LCM over here and we will write x - 2 + 1 over x - 2 plus square root of x ^ 2 - 4x + 4 is whole square of x - 2.

[00:01:46]Then we have + 2 over square root of x ^ 2 - 2x + 1 is whole square of x - 1.

[00:01:59]Then we have plus 2 again.

[00:02:02]equal to 0.

[00:02:05]So, our equation will become x - 1 over x - 2 plus square root of x - 2 whole square plus 2 over square root of x - 1 whole square plus 2 will be equal to 0.

[00:02:31]Now, we are going to use substitution.

[00:02:34]Let us write here x - 2 equal to y.

[00:02:42]So, our substitution is x - 2 equal to y. From this equation if I will write x - 1, then this will be equal to y + 1.

[00:02:54]Now, our equation will change into variable y. We will get y + 1 over y plus square root of y ^ 2 + 2 over square root of y + 1 whole square plus 2 will be equal to 0.

[00:03:21]Now, we will expand y + 1 whole square and we can write y + 1 over y plus square root of y ^ 2 + 2 over square root of y ^ 2 plus 2y plus 1 plus 2, so we can write 3 equal to 0.

[00:03:48]Now, we are going to use another substitution. Let us say square root of y ^ 2 + 2y + 3 This is equal to a.

[00:04:00]And our numerator square root y ^ 2 + 2, suppose this is equal to b.

[00:04:07]So, I will write our equation here.

[00:04:12]Now, we will square here and write a square value will be equal to y ^ 2 plus 2y plus 3.

[00:04:25]And if I will write b square after squaring this equation we'll get y ^ 2 plus 2.

[00:04:35]Now, we are going to subtract b square equation from a square. So, we have to change sign over here.

[00:04:43]We'll get a square minus b square in LHS.

[00:04:48]And in RHS, we'll write 2y plus 1.

[00:04:54]Now, this is a square minus b square in terms of y. We will find y value from here.

[00:05:01]So, I will subtract 1 from both the sides. We'll get a square minus b square minus 1 will be equal to 2y.

[00:05:10]Now, we have to divide by 2 both sides to get the value of y, which will be equal to a square minus b square minus 1 over 2.

[00:05:22]Now, as per our equation, y + 1 is also required, so let us find y + 1 directly from here.

[00:05:30]We'll add 1 to both sides. So, I will write a square minus b square minus 1 over 2 plus 1.

[00:05:39]This is y + 1 value. We have y.

[00:05:43]Now, we are calculating y + 1 will be equal to a square minus b square minus 1 plus 2 is plus 1 over 2.

[00:05:55]Now, we will convert our equation in variable a and b.

[00:06:01]We'll write here a square minus b square plus 1 divided by 2, which is y + 1 divided by y y is a square minus b square minus 1 divided by 2 plus square root y ^ 2 + 2 is b square root y ^ 2 + 2y + 3 is equal to a equal to 0.

[00:06:37]Now, we will cancel denominator 2 from here and we will write a square minus b square plus 1 over a square minus b square minus 1 plus b over a equal to 0.

[00:06:57]Now, we have to take LCM. So, we can write here a times a square minus b square plus 1 plus b times a square minus b square minus 1 will be equal to 0.

[00:07:21]Now, we are going to expand with respect to a square minus b square.

[00:07:27]So, let me write this equation here.

[00:07:33]Let's expand. We will get a times a square minus b square.

[00:07:42]Then we have a times 1, which we will write a plus b times a square minus b square.

[00:07:52]Then we have b times minus 1, we'll write minus b equal to 0.

[00:07:59]Now, we will club these two terms all together and we will club these two terms a and minus b. So, we can write a times a square minus b square plus b times a square minus b square plus a minus b equal to 0.

[00:08:25]Now, from these two terms, we can take a square minus b square common.

[00:08:31]So, I will write here a square minus b square is common.

[00:08:37]In other bracket, we need to write a plus b.

[00:08:42]Then we have a minus b over here equal to 0.

[00:08:47]We'll use difference of two squares identity here and we can write a plus b times a minus b times a plus b plus a minus b equal to 0.

[00:09:07]Now, from these two terms actually, we will take a a minus b common out.

[00:09:14]And we will write here a minus b is common.

[00:09:19]In other bracket, we can write a plus b times a plus b, which is a plus b whole square then we'll write plus 1 equal to 0.

[00:09:32]So, we'll use product zero rule and we can write a minus b This will be equal to 0 or a plus b whole square plus 1 equal to 0.

[00:09:50]Now, a plus b whole square will be always positive or equal to 0.

[00:09:58]So, our LHS must be greater than or equal to one.

[00:10:03]It cannot be equal to zero.

[00:10:06]So, LHS must be greater than or equal to one. So, we won't get any real solutions from this equation.

[00:10:14]So, the only equation which we are going to accept is a minus b equal to zero.

[00:10:20]So, we can write a equal to b.

[00:10:24]Now, a and b was our substitution.

[00:10:28]If I will write here square root of y square plus 2y plus 3 this was equal to a.

[00:10:39]And if I will write b substitution, then a square root of y square plus 2 this was equal to b.

[00:10:50]Now, we will put a equal to b.

[00:10:53]That means a square will be equal to b square.

[00:10:58]So, a square will be equal to y square plus 2y plus 3 will be equal to b square will be y square plus 2.

[00:11:10]Now, we will subtract y square from both the sides.

[00:11:15]So, we will get 2y equal to 2 minus 3 minus 1.

[00:11:21]Or we can write y will be equal to minus 1 over 2 after dividing both sides two.

[00:11:31]Now, y is our substitution x minus 2.

[00:11:34]So, we have to write here x minus 2 equal to minus half.

[00:11:39]x will be equal to 2 minus 1 over 2 which will be equal to 3 over 2.

[00:11:49]So, the only real solution for our equation would be 3 over 2.

[00:11:56]Let's verify our answer.

[00:11:59]Equation is written here.

[00:12:01]x is 3 by 2. Let me write here check.

[00:12:07]We will begin with left hand side.

[00:12:10]So, let us put x equal to 3 by 2. We will get LHS 1 plus a square root of x square will be 9 over 4 minus 4x will be minus 4 times 3 over 2 which will be minus 6.

[00:12:31]Then, we have plus 6.

[00:12:34]over a square root of x square we'll write 9 over 4 once again.

[00:12:40]minus 2x will be minus 3.

[00:12:43]Then, we have plus 3.

[00:12:46]This is our LHS.

[00:12:49]Plus 6 and minus 6 will get over. Plus 3 minus 3 will be zero.

[00:12:54]So, we are going to write LHS equal to 1 plus a square root of 9 over 4 divided by a square root of 9 over 4 which will get cancelled out. We'll get one here. So, LHS will become 1 plus 1 which is equal to 2.

[00:13:15]LHS is 2.

[00:13:18]Now, we will find RHS.

[00:13:21]RHS is 1 over 2 minus x. So, 1 over 2 minus 3 over 2 2 minus 3 over 2 is half.

[00:13:33]Denominator is half. We will flip it once we are taking reciprocal. So, it is 2 over 1 or we can write 2. LHS equal to RHS.

[00:13:45]Hence, our solution 3 over 2 is true and verified. I hope, friends, you will like this video. Thank you so very much for watching. Do not forget to like, share, and subscribe. Bye-bye. Till next video.

[00:13:57]Good luck.

[00:13:58]Take care. Bye-bye.