A Nice Algebra Problem | Math Olympiad x=?

Added: 2026-05-20

[00:00:00]Hello everyone, welcome to how to solve this very nice radical equation. The square root of X over 4 + 1 + the square root of X over 6 - 1 equal to the square root of X over 3 + 5. Our job is to find all possible values of X such that X is a real number. So, let's start. Since the LCM of this 4, 6, and 3 LCM of 4, 6, and 3 is equal to 12.

[00:00:35]So, we suppose that let X is equal to 12 * Y. So, this original equation will become the square root of 12 * Y over 4 + 1 + the square root of 12 * Y over 6 - 1 equal to the square root of 12 * Y over 3 + 5.

[00:01:19]Now, this is 4 one time and 4 three times. This is 6 one time, 6 two times and 3 one time, 3 four times. So, the equation will become the square root of 3 * Y + 1 + square root of 2 * Y - 1 equal to square root of 4 * Y + 5.

[00:01:51]Now, for real solution with this 3 Y + 1 3 * Y + 1 must be greater than or equal to zero.

[00:02:00]This implies that y must be greater than or equal to -1/3.

[00:02:07]And this 2y - 1 2y - 1 must be greater than or equal to zero.

[00:02:13]This implies that y must be greater than or equal to 1/2.

[00:02:19]This 4 times y + 5 must be greater than or equal to zero.

[00:02:26]This implies that y must be greater than or equal to -5/4.

[00:02:33]We combine these three, we get the domain of the equation y must be greater than or equal to 1/2.

[00:02:46]Now, we take square of both sides of this equation. So, this will become the square root of 3 times y + 1 plus the square root of 2 times y - 1 whole squared equal to the square root of 4 times y + 5 whole squared.

[00:03:21]Now, this square will be cancelled out with the square root by using this algebraic identity a + b whole squared equal to a squared plus 2 times a times b plus b squared.

[00:03:39]This will become square root of 3 times y + 1 whole squared plus 2 times the square root of We can write the product of these two terms uh under one square root. So, this will become three times y plus one times two times y minus one plus uh the square root of uh two times y minus one whole squared.

[00:04:21]And this is equal to The square will be cancelled out at right hand side. We're left with four times y plus uh five.

[00:04:34]Now, this square will be cancelled out with the square root and this square will be cancelled out with the square root. We are left with uh three times y plus one and from here plus two times y minus one plus two times the square root of This uh 3y times 2y will become 6y squared.

[00:05:00]And 3y times negative one negative 3y one times 2y plus two y and one times negative one negative one equal to four times y plus five.

[00:05:19]Now, this uh negative one will be cancelled out with this positive one.

[00:05:24]And we are left with uh This uh 3y plus 2y will become five times y plus two times the square root of Further simplify, this will become 6y squared.

[00:05:42]Negative 3y plus 2y negative y and this negative one equal to four times Y plus five.

[00:05:56]Move this five Y to the right hand side.

[00:05:58]This will become two times the square root of uh six Y squared minus Y minus one equal to four times Y plus five minus five Y.

[00:06:15]Further simplify, this will become two times square root of six Y squared minus Y minus one equal to five and four Y minus five Y will become negative Y.

[00:06:34]And from here, we take square of both sides.

[00:06:40]This square will be canceled with the square root and this the left hand side will become four times uh six Y squared minus Y minus one equal to This is five minus Y whole square will become five squared is 25 minus two times five times Y plus Y squared.

[00:07:14]Distribute this four four times the six Y squared 20 four Y squared.

[00:07:22]And the four times negative Y will become negative four Y. Four times negative one negative four equal to 25 negative two times five negative 10 Y plus this uh Y squared.

[00:07:41]Move all these three terms to the left-hand side. This will become 20 4 y squared minus y squared minus 4 * y + 10 * y minus 4 minus 25 equal to 0. Further simplify, this will become 23 y squared plus 6 * y minus 29 equal to 0.

[00:08:18]Now, this is a quadratic equation and is factorable. We write this 23 y squared break this is 6 y into two terms negative 23 y plus 29 y minus 29 equal to 0. From these two terms, we can factor out a 23 y. In bracket left, y minus 1. And from these two terms, we can factor out plus 29 in bracket left y minus 1 equal to 0.

[00:08:54]This y minus 1 is common, so we factor out this y minus 1. And in bracket left, 23 * y + 29 equal to 0.

[00:09:12]Either this expression y minus 1 equal to 0 or this expression 23 * y + 29 equal to 0.

[00:09:25]From this equation y will be equal to 1.

[00:09:30]And from this equation y will be equal to negative 29 over 23.

[00:09:39]Now, we remove any invalid solution for Y using the domain of the equation.

[00:09:45]The domain for Y is uh Domain for Y is uh Y must be greater than or equal to 1 over 2.

[00:10:05]The domain is Y must be greater than or equal to 1 over 2.

[00:10:11]Since this -29 over 23 is less than 1 over 2, so this will be rejected, and we accept only this value of Y, 1.

[00:10:23]Now, to find the values of X, recall that we have supposed Suppose that X is equal to 12 * Y.

[00:10:47]I suppose that X is equal to 12 * Y. Since Y is 1, so X will be equal to 12 * 1.

[00:11:00]And this implies that X is equal to 12.

[00:11:04]This is the final value of X. This is the only solution to this equation.