An Amazing Radical Equation | Can You Solve This?

Added: 2026-05-10

[00:00:01]Let's get it started is straight away with our substitution. Assume x square minus x minus 13 equal to a.

[00:00:11]Then x square minus x minus 12 will become a plus 1.

[00:00:17]And radicand x square minus x minus 11 will be a plus 2.

[00:00:25]So we can write our equation in variable a now. Cube root of a plus cube root of a plus 1 plus cube root of a plus 2 equal to 0.

[00:00:48]Now we know one formula from algebra.

[00:00:51]If cube root of a equal to p cube root of a plus 1 equal to q and cube root of a plus 2 equal to r if this sum equal to 0, then we know that p cube plus q cube plus r cube will be equal to three times p times q times r.

[00:01:23]So we are going to use this algebraic formula for our equation.

[00:01:30]So let me write here cube root of a equal to p.

[00:01:36]And after cubing both sides, I will get p cube value a.

[00:01:42]Similarly, cube root of a plus 1 if this is equal to q then by cubing both sides, I will find q cube equal to a plus 1.

[00:01:57]Similarly, we have substitution cube root of a plus two equal to r.

[00:02:05]Then by cubing both sides, we'll find the value of r cube, which will be equal to a plus two.

[00:02:13]Now, our target is to find p cube plus q cube plus r cube.

[00:02:19]So, we need to add these three equations. Let's add to find the value of p cube plus q cube plus r cube equal to a plus a plus a, 3a.

[00:02:38]1 plus 2, 3.

[00:02:41]So, we have p cube plus q cube plus r cube in terms of a.

[00:02:47]And we have pqr here.

[00:02:52]So, let's plug in all the values to get the equation in variable a.

[00:02:59]So, I will write here p cube plus q cube plus r cube equal to three times pqr.

[00:03:11]Now, we have p cube plus q cube plus r cube 3a plus 3.

[00:03:18]In RHS, we'll write three times p is cube root a.

[00:03:24]Q is cube root a plus one.

[00:03:29]And r is cube root of a plus two.

[00:03:35]We can club these three cube roots all together, and we can take three common from here.

[00:03:42]We will get three times a plus one equal to three times cube root of a times a plus one times a plus two.

[00:04:00]Now we can divide both sides by three.

[00:04:03]So three over three will be one.

[00:04:07]We will get LHS a plus one.

[00:04:11]Now RHS will be cube root of a times a plus one times a plus two.

[00:04:21]We have cube root in right hand side.

[00:04:24]To eliminate this cube root, we need to consider cubing both sides.

[00:04:31]Now we can cancel cube root with cube in right hand side.

[00:04:37]So we will get a plus one whole cube equal to a times a plus one times a plus two.

[00:04:52]Now we'll write all the terms to LHS.

[00:04:56]a plus one whole cube minus a times a plus one times a plus two equal to zero.

[00:05:09]Now we can take a plus one common from left hand side and we can write a plus one in other bracket, we'll get a plus one whole square minus of a times a plus two equal to zero.

[00:05:33]Now we have to expand a plus one whole square.

[00:05:37]So we'll write a plus one times a square plus two a plus one minus a times a minus a is - a * 2 - 2a = 0 Now + a 2 - a 2 + 2a - 2a will get over.

[00:06:03]We will write a + 1 * 1 = 0 or we will get a + 1 = 0 Now we have to subtract 1 from both the sides to get the value of a. So, a will be equal to -1.

[00:06:26]And a was our radicand x 2 - x - 13.

[00:06:33]So, in this place we can write x 2 - x - 13 = -1.

[00:06:43]Now we'll add 1 to both sides.

[00:06:47]And we will write x 2 - x - 13 + 1 -1 + 1 which will be equal to 0.

[00:07:00]-13 + 1 is -12. We will get x 2 - x - 12 = 0. So, we have one quadratic equation which we can solve using method of factorization.

[00:07:18]Now we have to split our middle term.

[00:07:21]So, I can write here x 2 -x can be written as -4x + 3x.

[00:07:30]If we'll multiply both, we're going to get -12x 2 -12 = 0. From first two terms, x is common. We'll get x - 4.

[00:07:43]From next two terms, we'll take three common and get x minus four equal to zero.

[00:07:51]Now, x minus four is overall common.

[00:07:54]So, we will get x minus four times x plus three equal to zero.

[00:08:04]Using product zero rule, we will write two equations either x minus four equal to zero or x plus three equal to zero.

[00:08:14]So, from our first equation, we are going to get our first real solution x equal to four.

[00:08:22]And from second equation, we'll be getting second real solution x equal to negative three.

[00:08:29]Now, we have to verify our solutions whether they will satisfy equation or not.

[00:08:35]So, let us write here check.

[00:08:40]Equation is written here.

[00:08:44]We will begin with first solution x equal to four.

[00:08:48]Let's write here LHS straightaway by putting x equal to four. We will get cube root of x is square, so we will write four is square minus x, so minus four minus 13 plus cube root of four is square minus four minus 12 plus cube root of four is square minus four minus 11.

[00:09:25]Now, four is square is 16. Minus four is 12. So, 12 minus 13 is minus one. We'll write here cube root of minus one plus four is square minus four is 12. 12 minus 12 is 0. We'll write here cube root 0.

[00:09:42]plus 16 minus 4 is 12. 12 minus 11 is 1.

[00:09:47]We'll write here cube root of 1.

[00:09:50]For real x, cube root of minus 1 is minus 1.

[00:09:55]Plus cube root 0 is 0.

[00:09:58]Plus cube root 1 is 1. We are going to get this addition 0, and 0 is nothing but our RHS.

[00:10:07]Hence, our solution x equal to 4 is true and verified.

[00:10:14]Now, the turn of second solution.

[00:10:17]We'll write here x equal to -3.

[00:10:21]Again, we will begin with left-hand side.

[00:10:26]We'll find cube root of x squared will be -3 squared. I'm writing 9.

[00:10:32]Minus x will be plus 3.

[00:10:35]Minus 13.

[00:10:38]Plus cube root of 9 plus 3 minus 12 plus cube root of 9 plus 3 minus 11.

[00:10:54]So, 9 plus 3 is 12. 12 minus 13 is minus 1. We'll get cube root of minus 1 plus 12 minus 12 is 0. Cube root of 0 plus cube root of 1.

[00:11:10]So, cube root of minus 1 we'll write minus 1.

[00:11:13]Here, we can write 0. 1. This addition will give us 0, and 0 is our right-hand side.

[00:11:21]LHS equal to RHS.

[00:11:23]Hence, our solution minus 3 is true and verified.

[00:11:29]So, we have two real values of x which will satisfy equation x equal to 4 and x equal to -3.

[00:11:36]I hope friends you will like this video.

[00:11:38]Thank you so very much for watching. Do not forget to like, share, subscribe.

[00:11:42]Bye-bye.