This approach prioritizes structural elegance over brute-force calculation, fostering the mathematical intuition necessary for advanced problem-solving. It effectively transforms a routine algebraic task into a sophisticated exercise in pattern recognition.
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Deep Dive
Don't Think About LCM Immediately..... Rather Do This!Added:
Hello friends, welcome back to MC Gyan.
Today in this video we have one very very interesting and challenging question from rational equations, which will be solving for the real values of X.
So, let's get it started by writing here denominators cannot be zero. So, X cannot be equal to 4 5 7 8.
Now, we are going to check the square of denominators.
Here we have X minus 4.
Its a square will be X is square minus 8X plus 16. Here we have 14.
If you will check a square of X minus 8, then it is X is square minus 16X plus 64. Here we have 62.
The square of X minus 5 is X is square minus 30X. Here we have 26.
But as per formula it is 25.
And X minus 7 whole is square will be X is square minus 14X plus 49. Here we have 50.
So, let's write our numerator in terms of denominator is square.
I will write the equation X is square minus 8X will write 14 as 16 minus 2.
In denominator will write X minus 4.
Similarly here will write X is square minus 16X plus 64 minus 2 over X minus 8.
equal to in RHS will write X is square minus 10X plus 25 plus 1 over X minus 5 plus X is square minus 14X plus 49 plus 1.
And in denominator will write X minus 7.
Now, we are going to pick first three terms of numerators.
These are perfect squares.
Let's write perfect is square of their respective denominators. We will write here X minus 4 whole is square minus 2 over X minus 4 plus X minus 8 whole is square minus 2 over X minus 8 equal to X minus 5 whole is square plus 1 over X minus 5 plus X minus 7 whole is square plus 1 over X minus 7.
Now, we'll split numerator with respect to their denominators.
So, we can write X minus 4 whole is square over X minus 4 X minus 4.
Then we have minus 2 over X minus 4 plus X minus 8 whole is square over X minus 8 is X minus 8 minus 2 over X minus 8 equal to X minus 5 plus 1 over X minus 5 plus X minus 7 plus 1 over X minus 7.
Now, we will add X minus 4 and X minus 8 from LHS.
We can write X plus X is 2X.
minus 4 minus 8 is minus 12.
From RHS we have X minus 5 and X minus 7 which will give us 2X minus 12 after adding them.
Now, remaining terms would be as it is.
We'll write minus 2 over X minus 4 minus 2 over X minus 8.
In RHS we'll write plus 1 over X minus 5 plus 1 over X minus 7.
Now, we'll subtract 2X minus 12 from both the sides.
So, 2X minus 12 will get over.
We will write our remaining equation minus 2 over X minus 4 minus 2 over X minus 8 equal to 1 over X minus 5 plus 1 over X minus 7.
Now, we'll write all the terms to right hand side.
So, we can write here 2 over X minus 4 plus 2 over X minus 8 plus 1 over X minus 5 plus 1 over X minus 7 equal to zero.
We can take two common from first two terms and we'll be writing 1 over X minus 4 plus 1 over X minus 8 plus 1 over X minus 5 plus 1 over X minus 7 equal to zero.
Now, we'll use substitution.
Let us say X minus 6 this is equal to U.
So, we'll write here X minus 4 in terms of U will be U plus 2.
And if you will write X minus 8 then we have to write U minus 2.
Similarly, we'll write X minus 5 will be equal to U plus 1.
And X minus 7 will be U minus 1.
So, our equation will become two times 1 over U plus 2 plus 1 over U minus 2 plus 1 over U plus 1 plus 1 over U minus 1 equal to zero.
Now, we are going to take LCM. LCM of these two brackets.
So, we will write here two times 1 over U plus 2 plus 1 over U minus 2 will write U plus 2 times U minus 2 in the denominator and we have to add in the numerator. So, U plus 2 plus U minus 2 plus in denominator we'll multiply both the denominator brackets U plus 1 times U minus 1 and in numerator we will add both U plus 1 plus U minus 1 equal to zero.
Now, we'll cancel plus 2 and minus 2 plus 1 and minus 1.
So, we will write two times in bracket we'll write 2U over A plus B times A minus We know that this is difference of two is square formula.
A is square minus B is square. So, we'll write here U is square minus 2 is square plus U plus U will be 2U.
In denominator we'll write U is square minus 1 is square equal to zero.
But here we'll write bracket only because of multiplier is two with only first term.
Now we will take 2U common from the numerator.
So, I will take two inside and we will take 2U outside. So, we will write 2U in bracket we'll write 2 over U is square minus 4 plus 1 over U is square minus 1 equal to zero.
Now, we'll use product zero rule.
So, we can write either U equal to zero or 2 over U is square minus 4 plus 1 over U is square minus 1 equal to zero.
From U equal to zero equation we have U value X minus 6. So, we will write here X minus 6 equal to zero.
So X will be equal to six. Our first real solution is X equal to six.
Now we have to solve this equation.
We'll take LCM once again.
So we can write Let me write equation here.
Two over U is square minus four plus one over U is square minus one equal to zero.
We'll multiply both the denominators in the denominator here. U is square minus four times U is square minus one.
And in numerator we will multiply two with U is square minus one.
And one with U is square minus four equal to zero.
So denominator cannot be zero. We'll write here two times U is square minus one.
Plus U is square minus four will be equal to zero. Or we can write two U is square minus two plus U is square minus four equal to zero.
We'll add two U is square with U is square three U is square minus two minus four minus six equal to zero.
We'll add six to both sides and write three U is square equal to six.
Now we have to divide by three both sides.
Three over three will be one.
And six over three will be two.
So we can write U is square equal to two.
After taking a square root both sides, we'll write U will be equal to plus minus the square root of Now U is X minus six.
So we will write U equal to plus minus the square root of two.
In place of U we'll write X minus six equal to plus minus the square root two.
After adding six to both sides, we are going to write here two solutions. X equal to six plus minus the square root.
So in total we have three real solutions.
X1 it is six.
X2 it is six plus root two.
And X3 will be equal to six minus root I hope friends you will like this video.
Thank you so very much for watching. Do not forget to like, share, subscribe.
Bye-bye till next video. Good luck.
Bye-bye.
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