The video provides a clear demonstration of using basic identities to solve radical systems, even if the "95% fail" claim is exaggerated. It’s a solid reminder that complex problems often have simple algebraic solutions.
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This System Stumps 95% of Students! | Math ChallengeAdded:
Hello friends, welcome back to NC John.
Today in this video we'll be solving one very interesting system of equation problem for the real values of X and Y.
So, let's get it started by writing here equation number one.
And this equation we will call equation number two.
Now, we are going to subtract second equation from first one.
So, we will write here 1 - 2.
So, we are going to get X - Y square root X square - Y square will get over will be equal to 5 - 3.
Or we can write X - Y will be equal to 2.
So, from this equation we will get Y in terms of X Y equal to X - 2.
Let's say this is our equation number three.
Now, we are going to use equation one and equation three.
We will write X + the square root of X square - Y square equal to 5.
This is our equation one.
Now, we'll use equation three and write X + the square root of X square Y is X - 2. We'll write X - 2 whole square.
Will be equal to 5.
Now, we'll be using A - B whole square identity over here.
So, let me write A - B whole square equal to A square 2 * AB + B square.
Which we'll apply here and we will get x plus the square root of x square minus x minus two whole square is x square minus four x plus four.
And RHS is five.
We'll write x plus the square root of x square minus x square plus four x minus four equal to five.
Now, we'll cancel x square with minus x square.
So, this will give us x plus the square root of four x minus four equal to five.
Now, we'll be subtracting x from both the sides.
So, we'll write minus x in RHS.
Plus and minus x will get over from left hand side. We will get the square root of four x minus four equal to five minus x.
Now, we have a square root in LHS. So, we are going to a square both sides.
But, before that, if we use basic algebra then we have two conditions from this equation if x has to be real. Condition number one radicand must be non-negative.
So, we will write here four x minus four should be greater than or equal to zero.
Condition number two, a square root must be non-negative.
So, five minus x equal to a square root.
So, this value must be greater than or equal to zero.
So, we'll write our second condition, 5 - x should be greater than or equal to zero.
So, from first inequality we will write 4x should be greater than or equal to 4.
Or we will write x should be greater than or equal to 1.
From left-hand side.
And from right-hand side, 5 - x is greater than or equal to zero. So, from here we will get x should be less than or equal to 5.
So, there are two conditions which x has to satisfy. So, we will get x should be in the interval 1 to 5.
Our x value, if it is in in the interval 1 to 5 then we'll say our solutions are real.
Now, we'll be squaring both sides.
We will write square root of 4x - 4 equal to 5 - x.
We'll put power two. We'll put power two.
Now, a square root and a square will get over from LHS.
And we'll be using a minus b whole square formula in RHS.
So, we can write 4x - 4 equal to 5 square is 25 minus 2ab, two times five times x, minus 10x plus x square.
Now, we'll take all the terms to RHS.
So, we will get x is square -10x -4x -14x 25 + 4 29 equal to zero.
Now, we have one quadratic equation. We can use quadratic formula.
So, I need to write here a value which is coefficient of x is square one.
B value which is coefficient of x -14 and c will be equal to constant 29.
As per formula, x will be equal to -b plus minus the square root of b square minus four times a times c over two times a.
Let's plug in all the values of a, b, and c. We will get minus of minus 14 +14 plus minus the square root of -14 is square.
It is 196.
minus four times one times 29 will be equal to 116 over two times one will be two.
So, we can write x will be equal to 14 plus minus the square root of 196 minus 116 80 over two.
Now, we can factor 80 as 16 times five.
So, x will be equal to 14 plus minus the square root of 16 times five over two.
Now, the square root of 16 four.
So, we will write 14 plus minus four times the square root of five over two.
Which will give us seven plus minus two times the square root five.
So, we have two x values.
Seven plus two square root five.
And second is seven minus two square root five.
Square root five is approximately 2.23.
So, two times 2.23 is 4.46.
So, 7 plus 4.46 approximately 11.46.
And 7 minus 4.46 will be approximately 2. 5 4.
Now, our condition on x is x should be in the interval one to five.
So, we are going to reject seven plus two square root five as it is greater than five.
So, we can accept 2.54 which is seven minus two square root five.
So, x will be equal to seven minus two square root five.
Now, we have to find y value.
And y is x minus two.
We'll put the value of x seven minus two square root five minus two.
So, 7 minus 2 is five. We are going to get y value five minus two square root five.
So, x is seven - 2 square root 5.
And Y is 5 - 2 square root 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.
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