Can You Crack This Nested Radical Equation?

[00:00:02]Hello friends, welcome to InfyGyan. In this video, we are going to solve one very interesting nested square root equation for the real values of x.

[00:00:12]So, let's get it started.

[00:00:14]We have x in left-hand side and a square root in right-hand side. So, from left-hand side, x must be non-negative.

[00:00:22]But, we have x in the denominator.

[00:00:25]So, x will be only positive number for x to be real.

[00:00:32]If you are thinking about a squaring both sides, let's check what will happen if we will a square both sides.

[00:00:40]We'll cancel a square root with a square from RHS.

[00:00:45]We'll write x a square equal to 2 + 1 over the square root of 2 + 1 over x.

[00:00:58]Now, we will subtract two from both the sides and get x a square minus two equal to 1 over the square root of 2 + 1 over x.

[00:01:13]Now, we have a square root in the denominator. We will a square once again to remove or eliminate the square root.

[00:01:21]So, we'll put whole a square in LHS.

[00:01:25]Whole a square in RHS.

[00:01:27]We will get x a square minus two whole a square equal to 1 over 2 + 1 over x.

[00:01:40]Which will become x over 2x + 1.

[00:01:47]Now, we will cross multiply. We'll be getting x a square minus two whole a square times 2x + 1 equal to x.

[00:02:00]So, we'll be having 1° 5 polynomial equation.

[00:02:04]And its roots won't be that easy.

[00:02:09]So, we will cancel this method and we will follow one conventional method.

[00:02:15]Here we have a square root 2 + 1 over x.

[00:02:21]If I will replace this denominator by x, it is satisfying our original equation.

[00:02:28]I will write x equal to the square root of 2 + 1 over x.

[00:02:38]Now, condition on x is x should be greater than zero.

[00:02:44]Now, we'll be squaring both sides.

[00:02:47]So, x squared will become 2 + 1 over x.

[00:02:55]Now, x cannot be zero. So, we'll multiply whole equation by x.

[00:03:02]So, I will write here times x.

[00:03:06]LHS will become x squared times x, x cubed equal to 2 times x, 2x + 1 over x times x, 1.

[00:03:19]We'll write all the terms to one side.

[00:03:22]x cubed - 2x - 1 equal to zero.

[00:03:30]Now, equation is cubic. We will use factorization method to find its roots.

[00:03:37]So, I will write x cubed - 2x can be written as - x - x.

[00:03:44]Then we have - 1 equal to zero.

[00:03:47]From first two terms, we can take x common.

[00:03:51]So, there will be x is squared minus one in the bracket.

[00:03:56]From next two terms, we will take minus one common out.

[00:04:02]So, in other bracket, we will get x plus one equal to zero.

[00:04:09]Now, we will use difference of two squares formula over here, a squared minus b squared.

[00:04:15]It is a plus b times a minus b minus one times x plus one.

[00:04:28]Now, we have x plus one overall common.

[00:04:33]So, we will get x plus one times x times x minus one minus one.

[00:04:41]We'll write here x plus one.

[00:04:44]In other bracket, we'll write x times x minus one minus one equal to zero.

[00:04:55]Or we can write x plus one times x is squared minus x minus one equal to zero.

[00:05:09]Now, we will use product zero rule.

[00:05:12]So, we will get two equations. Either x plus one equal to zero or x is squared minus x minus one equal to zero.

[00:05:25]So, from first equation, x plus one equal to zero, we will subtract one from both the sides.

[00:05:33]And we'll cancel plus one minus one from LHS. Our x will become negative one.

[00:05:40]As per condition on real numbers, our x should be greater than zero.

[00:05:47]So, we are not going to accept negative one as real solution.

[00:05:52]Now, we have one quadratic equation only.

[00:05:57]Which we will solve using quadratic formula method.

[00:06:01]So, if I will write equation over here, x ^ 2 - x - 1 = 0.

[00:06:08]A is coefficient of x ^ 2 1. B is coefficient of x negative one. And c is constant negative one.

[00:06:20]We'll write here quadratic formula, x = - b plus minus the square root of b ^ 2 - 4ac over two times a.

[00:06:35]Let's apply all the values over here. We will get minus of minus one plus one.

[00:06:43]Plus minus the square root of b ^ 2 - 1 ^ 2 + 1 minus four times one times minus one.

[00:06:57]And in denominator two times one is two.

[00:07:03]So, we will get one plus minus the square root of one plus four over two.

[00:07:14]One plus four is five. So, we will write here one plus minus the square root of five over two.

[00:07:24]So, we have two values of x. Let me write separately x = one plus the square root five over two.

[00:07:37]And with minus sign in between one minus the square root five over two.

[00:07:43]Now, we know that the square root 5 is approximately 2.23.

[00:07:50]1 - 2.23, number will be negative.

[00:07:55]So, we can cancel 1 - the square root 5 over 2.

[00:08:00]So, the accepted x would be x equal to 1 + the square root 5 over 2, the only real solution for our equation.

[00:08:13]I hope, friends, you will like this video. Thank you so very much for watching.

[00:08:19]Do not forget to like, share, and subscribe.

[00:08:21]Bye-bye till next video. Good luck. Take care. Goodbye.