Outstanding Radical Equation | Can You Solve This?

Welcome to InfiGyaan. Today in this video we have one very interesting question from radical equations which we will be solving for real values of x. So let's get started by multiplying 3 in the numerator denominator. So we will write times 3 in the denominator also. Here we will multiply with 3. So in the denominator we will multiply by 3. Now we have to expand simplify and we will get fifth root of 5 times 3 is 15. So 15x plus 6 times 3 is 18 over 81 times 3 is 243 x plus 243 plus fifth root of 18x plus 15 over 243x plus 243 equal to 1. Now we will take 243 common from the denominator. So we will get fifth root of 243 in the denominator which we can further write fifth root of 3 power 5. 3 power 5 is 243. So fifth root and power 5 will be over. We are going to get 3 in the denominator. Once we will take 243 common from the radicand denominator. So I will write 1 over 3 fifth root of 15x plus 18 over x plus 1 plus 1 over 3 fifth root of 18x plus 15 over x plus 1 equal to 1. Now we will multiply this equation by 3 both sides. So let me write here times 3. So in the RHS we will write times 3. So 3 and denominator 3 will be over from left hand side. We will get our equation fifth root of 15x plus 18 over x plus 1 plus fifth root of 18x plus 15 over x plus 1 will be equal to 3. Now we have to solve this equation. Let me write this equation here. Here it is. Now we will use substitution. Let us say this radicand 15x plus 18 over x plus 1 equal to a power 5 and 18x plus 15 over x plus 1 we will consider b power 5. So we can write our first equation fifth root of a power 5 plus fifth root of b power 5 equal to 3. Fifth root power 5 will be over. We will write equation number 1 a plus b equal to 3. Now we will add a power 5 with b power 5. So we will be writing here 15x plus 18 and 18x plus 15 in the numerator with common denominator x plus 1. 15 plus 18 is 33 x. 18 plus 15 is 33 over x plus 1. Now we will take 33 common out from the numerator. We will get x plus 1 over x plus 1 which is 1. So our second equation is also ready a power 5 plus b power 5 equal to 33. This is our equation number 2. Now we will use here algebraic identity. Let me write both the equations a plus b equal to 3. a power 5 plus b power 5 equal to 33. Now I will calculate a square plus b square using algebraic formula which is a plus b whole square minus 2 times a b. Now better is to assume a b equal to u suppose u. So we will write here a plus b whole square is 3 square minus 2 times a b. a b is u. So we will write 9 minus 2 u. This is our equation number 3. Now we are going to calculate a cube plus b cube using algebraic identity. We can write a plus b whole cube minus 3 a b times a plus b. Now we have a plus b 3. So we will write here 3 cube minus 3 times u a plus b is 3. So I can write will be equal to 3 cube minus 9 u. Let us take 3 common out. So there will be 3 square left. I will write 9 minus 3. This we will call equation number 4. Now we are going to multiply 3 with 4. So I will write here 3 times 4. 3 was a square plus b square in the LHS. Then we will write a cube plus b cube. Now we have to multiply both the right hand side. So we will get 9 minus 2 u times 3 into 9 minus 3 u. Now we will multiply a square with a cube a power 5 b square with b cube b power 5 a square b cube plus b square a cube from we will take a square b square common out. So in the bracket we can write a plus b equal to 3 times 9 minus 2 u times 9 minus 3 u. Now a power 5 plus b power 5 equation number 2 we can write 33. a b was u so we will write u square here. a plus b was equation 1 value was 3 equal to 3 times 9 minus 2 u times 9 minus 3 u. Now our equation is divisible by 3. So let us divide both sides by 3. Now we will cancel 33. Here also we will write 11 plus u square equal to 9 minus 2 u times 9 minus 3 u. Now we have to simplify our RHS. So I will get here 9 times 9 is 81 minus 27 u minus 18 u minus 45 u plus 6 u square. LHS is 11 plus u square. Now take all the terms to right hand side. We will get 5 u square minus 45 u 81 minus 11 is 70 equal to 0. Now equation is divisible by 5. We will divide both sides by 5 u square minus 9 u plus 14 equal to 0 which we can easily factorize in u minus 2 times u minus 7 equal to 0. Now we will use product 0 rule. So either u minus 2 equal to 0 we are going to get u equal to 2 or u minus 7 equal to 0 we are going to get u equal to 7. Now this is the value of ab. u was our substitution ab. So I will write here ab equal to 2, ab equal to 7. Let me write here equation 1 also. a plus b equal to 3. From this equation I can write the value of b in terms of a 3 minus a which we are going to plug in here. So I can write a times 3 minus a equal to 2 7 or 3 a minus a square equal to 2 and 7. Now we will write two equations from here. Equation 1. Take all the terms to RHS. We will get a square minus 3a plus 2 equal to 0 and second equation would be a square minus 3a plus 7 equal to 0. Now we will start with second equation and check discriminant only which is b square minus 4 times ac. So here we will write minus 3 square minus 4 times 1 times 7 minus 3 square is 9 4 times 7 is 28. We are going to get minus value minus 19 negative discriminant is negative we are going to get complex solutions, complex roots or solutions. So we can reject this quadratic as we are looking for real solutions. Now we have a square minus 3a plus 2 equal to 0 equation which we can easily factorize a minus 1 times a minus 2 equal to 0. Using product zero rule we will get a equal to 1, a equal to 2. So final conclusion is a is 1, a is 2. Now a was our substitution. Let me write here a equal to 1, a equal to 2 and our substitution was 15x plus 18 over x plus 1. This was equal to a power 5. So we will calculate a power 5 from here which will be equal to 1 power 5, 2 power 5. So a power 5 will be 1 and 32. Now we have a power 5, 15x plus 18 over x plus 1. Let us write 15x plus 18 over x plus 1 equal to 1 and 15x plus 18 over x plus 1 equal to 32. Now we will cross multiply. 15x plus 18 equal to x plus 1. As per our equation denominator cannot be 0, so x cannot be negative 1. So x plus 1, now we will take variable to one side. So 14x minus x, 14x and in RHS 1 minus 18, we will write minus 17. Now we have to divide by 14 both sides. So our first real answer or solution is minus 17 over 14. Let me write here x1. Now we will get x2 from the second equation. Let's cross multiply. We will write here 15x plus 18 equal to 32x plus 32. Take 15x to the RHS, 32 to the LHS. So we will get 18 minus 32 equal to 32x minus 15x. So 18 minus 32 is minus 14 equal to 32 minus 15 is 17x. Now we have to divide by 17 both sides to get our second solution. So I will write here x2. x2 will be equal to minus 14 over 17. So we have two real values, two real solutions minus 17 over 14 and minus 14 over 17. 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.