3 Simple Steps to Solve This Radical Equation with Ease!

Hello everyone. Welcome back to NPGyan. Today in this video we have one very very interesting question from radical equations which will be solving for the real values of x. Here x cannot be negative 1. If we will pass x equal to negative 1, numerator will become 0, denominator will become 0. So 0 by 0 form is indeterminate form. We cannot accept it. So let's get started by taking 81 common from this radical. So we will write here 5th root of 81 times 5th root of x plus 1 over 5th root of 9x plus 2 plus 5th root of 2x plus 9 equal to 1. Now we will use one algebraic manipulation step. We are going to multiply our numerator and denominator by 5th root of 3. So that 3 times 81 will become 243 which is 3 power 5. So 5th root of 3 power 5 would be 3. Here we will write 5th root of 3. The bracket will multiply. Now we are going to multiply these two radicands. 3 with 81 will be getting 243. So I can write here 5th root of 243 times 5th root of x plus 1 over. We are going to multiply 5th root of 3 with both the terms. So we can get 5th root of 9 times 3, 27x. 3 times 2, 6 plus 5th root of 3 times 2, 6x plus 3 times 9, 27 equal to 1. Now 5th root of 243 is 3. Let me write here 5th root of 243 will be equal to 5th root of 3 power 5. So 5th root and power 5 will be over. We will get 3. So I can write here 3 times 5th root of x plus 1 over 5th root of 27x plus 6 plus 5th root of 6x plus 27 equal to 1. Now we will divide both sides by 3. So let me write here divided by 3. So our numerator 3 will be over and in RHS we need to write 3 in the denominator. So I can write our equation 5th root of x plus 1 in the numerator and in the denominator we are going to write 5th root of 27x plus 6 plus 5th root of 6x plus 27 equal to 1 over 3. Now we are going to flip numerator and denominator. To consider reciprocal both sides we will get our equation 5th root of 27x plus 6 plus 5th root of 6x plus 27 over 5th root of x plus 1 and reciprocal of 1 over 3 is 3. Now we will split numerator with respect to denominator and we can write 5th root of 27x plus 6 over x plus 1 plus 5th root of 6x plus 27 over x plus 1 equal to 3. Now we will use substitutions. Let us say first radicand 27x plus 6 over x plus 1 equal to a power 5 and 6x plus 27 over x plus 1 b power 5. So we can write our first equation 5th root of a power 5 which is a plus 5th root of b power 5 b equal to 3. This equation is equation number 1. Now we will add a power 5 and b power 5. Let us check what will come out. So we will write our common denominator. We will write a power 5 plus b power 5 equal to common denominator is x plus 1. Now we will add numerator terms 27x plus 6 plus 6x plus 27. So we will get 27 plus 6, 33x. 6 plus 27, 33 over x plus 1. So we can take 33 common out. In the bracket we are going to write x plus 1 which is match with our denominator. Now we will get a power 5 plus b power 5 equal to 33. This equation we will call equation 2. Now I will use binomial expansion. Let me write our first equation also a plus b equal to 3. Now I will use the direct formula for a plus b whole power 5. It is equal to a power 5 plus b power 5 plus 5 times ab in the bracket a cube plus b cube plus 10 times a square b square times a plus b. Now we have to calculate a cube plus b cube. We can plug in here a plus b value which is 3. Here also we will put 3 but a cube plus b cube is not known. So I will write here a plus b equal to 3 equation and we will consider cubing both sides. Now we can apply a plus b whole cube in LHS and write a cube plus b cube plus 3ab times a plus b equal to 3 cube 27. Now a plus b is 3 so here we are going to put 3. So we will get a cube plus b cube plus 9ab equal to 27. Let me write here a cube plus b cube plus 9ab equal to 27 or we can write a cube plus b cube value equal to 27 minus 9 times ab. Now I can take 3 common out. So in the bracket I can write 9 minus 3ab. Now I will write our formula once again. Here it is. Now we will put a plus b equal to 3. So LHS will become 3 power 5 equal to a power 5 plus b power 5 equation number 2. We will write 33 plus 5ab times a cube plus b cube which is written here 3 times in the bracket 9 minus 3ab. So we will write 3 times 9 minus 3ab. Then we have a plus b this value equal to 3. So we can write 10 times 3 30 a square b square. Now our equation is divisible by 3. So we will divide both sides by 3. Let's divide divided by 3. So LHS will become 3 power 4. RHS will be 33 over 3 is 11. Then we will write plus 5 times ab in the bracket 9 minus 3ab. 30 over 3 is 10. So we will write 10 times a square b square. Now 3 power 4 is 81 and if we will take 11 to the LHS 81 minus 11 is 70 equal to 5ab times 9. So I can write 45ab 5ab times minus 3ab. So minus 15 a square b square plus 10 a square b square. So we can write minus 5 a square b square. Now this equation is divisible by 5 and we will take all the terms to one side. We will get quadratic in ab. If I will write directly a square b square minus 9ab plus 14 equal to 0. Now we have to solve this quadratic equation which is in ab for the value of ab. So I can write quadratic as ab whole a square minus 9 times ab plus 14 equal to 0. Now we can see the factors of 14 as 2 times 7 addition of both will give us middle term coefficient 9. I will write that factors ab minus 2 times ab minus 7 equal to 0. Using product zero rule we will be getting ab equal to 2 ab equal to 7. Now I will write our first equation a plus b equal to 3. From here I can write b equal to 3 minus a which we will apply here. So I can write a times 3 minus a equal to 2 equal to 7 or 3a minus a square equal to 2 and 7. Now we will frame two quadratic equations one with 2 another with 7. So I can write a square minus 3a plus 2 equal to 0. This is our first quadratic and second equation would be a square minus 3a plus 7 equal to 0. Now I will check discriminant for second quadratic equation. So I will write b equal to b square minus 4ac. So minus 3 square minus 4 times 1 times 7 minus 3 square is 9 minus 4 times 7 is 28. Value is coming out negative 19. Discriminant is negative. We are going to get complex solutions. So this quadratic equation we are going to reject. So we have to solve a square minus 3a plus 2 equal to 0 for our real calculations real solution. So I will write a square minus 3a plus 2 equal to 0. Again we can write here factors of 2 as 2 times 1 and addition of both will give us middle term coefficient 3. So splitting is very easy. We will write and get a minus 1 times a minus 2 equal to 0. Using product zero rule we will get either a minus 1 equal to 0 or a minus 2 equal to 0. So we will be getting a equal to 1 a equal to 2. Now if you remember our substitution was a power 5 equal to 27x plus 6 over x plus 1. So I need to calculate a power 5 from here. So I will write here 1 power 5. Here we will be writing 2 power 5. So the value of a power 5 will be 1 and 32. Now we have to replace a power 5 with 27x plus 6 over x plus 1. So let us write 27x plus 6 over x plus 1 equal to 1 and 27x plus 6 over x plus 1 equal to 32. Now we will cross multiply. So I will write equation here 27x plus 6 equal to 1 times x plus 1. So we can write x plus 1. Now we will take x to the LHS and 6 to the RHS. So I can write 27x minus x equal to 1 minus 6. 26x will be equal to negative 5. Now we have to only divide by 26 both sides. We are going to get our first real solution. So x will become negative 5 over 26. Now we will consider our second equation 27x plus 6 over x plus 1 equal to 32. Cross multiply. We will get 27x plus 6 equal to 32x plus 32. Now we will take 6 to the RHS, 32x to the LHS. So 27 minus 32 will give us minus 5x in the LHS and RHS would be 32 minus 6, 26. Now we have to divide by minus 5 both sides to collect our second real solution. So I will write here x equal to minus 5 over 5. So we have in total two real solutions. If I will write here x equal to minus 5 over 26, x equal to minus 26 over 5. 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. Good luck. Take care. Bye-bye.