To solve equations containing radicals like x^(1/2), substitute a new variable for the radical expression, simplify the equation using exponent rules, and then back-substitute to find the original variable. For example, in x³ + x^(1/2) = 66, let y = x^(1/2), so x = y², transforming the equation to y^6 + y = 66, which can be solved by recognizing that 64 + 2 = 66, giving y = 2 and thus x = 4.
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Olympiad Mathematics | Germany | Can You Solve This?Added:
Hi, everyone.
You're welcome.
Let's provide the solution to this equation here.
Okay, so we have x to the power of 3 plus x to the power of 1 over 2 equals 66.
Now, how do you solve this problem?
Now, let me pick out what we have here, right?
Now, we have x to be what? x to the power of 1 over 2.
Let's this be equal to y.
Right?
So, that anywhere we find x to the power of 1 over 2, we're going to write y.
Then, how do we do this? What do we do to this one here?
Now, from here we have to remove the power of 1 over 2 by squaring it, right?
And on the other hand, we're going to square the y.
This will take this out.
And at the end of the day x will be equal to y squared, right?
Okay. So, if this is true then we can put these values you know, this y and y squared into the equation.
Permit me to write the equation again so that we can see it here.
x to the power of 3 plus x to the power of 1 over 2 equals 66.
Now, this is x to the power of 3, but we now know that our x is y squared, right?
So, that means that in place of x cubed I'm going to write y squared.
Then, there's power of 3.
Plus, look at that.
Um y to the power x to the power of 1 over 2 is Y from there.
So, this is Y and it is all equal to 66.
We will not stop here, right?
So, to continue with this, we know the relationship between these two powers because A to the power of M to the power of N can be written as A to the power MN.
Right? Because what we have done is just to multiply by M by N.
Okay, let me remove this for the sake of space.
Now, we are going to have Y to the power of 6 plus Y equals 66.
Now, we have to look at what we have on the right because definitely we cannot add these two.
But then, from here, let's bring out a number from 66 that can be expressed to have the power of 6.
Okay, I already have a number.
Remember, you can bring out 40, you can bring out 60, but you cannot express them to the power of 6.
So, what do we do?
For Okay, let me write this first. Y's Y's Y to the power of 6 plus Y equals 64 plus two.
64 plus two. Oh, by the way, we are getting only the complete solution I mean the real solution.
We are getting only the real solution.
So, we will not have to go too far.
Okay. So, from here now we have Y Y to the power of 6 plus Y is equal to look at 64.
To express 64 in the power of six, you're going to write two to the power of six.
Then we have plus two.
So, at this point, we can begin to compare.
Okay, so by comparison, we're going to say um y to the power of six should be equal to two to the power of six.
And now that the powers are the same, we can equate the bases.
So, that y will be equal to two.
Right?
And then or if you compare this and this, you're going to get y to be equal to two directly.
But we will not stop there, right?
Because we are getting the value of x.
And at some point, we said that let x to the power of 1 over 2 be equal to y.
Okay?
So, if this is true, then we can say that You know, the value of y is two now, so we can say x to the power of 1 over 2 is equal to two.
And to remove that power from there, we will have to square both sides of the equation.
And then x will be equal to two squared, which is Right? So, this right now becomes the value of x that will satisfy the given equation, which is x to the power of three plus um we have um x to the power of 1 over 2 equals 66.
This is the equation, right? We want to verify.
If you put in the value of x to be four, then you have four cubed plus four to the power of one over two.
Four cubed is 64.
Then four to the power of one over two is is two.
Cuz this is the same as the square root of four.
So at the end of the day we have six plus four, which will give us 66.
So the value of x to be equal to four truly satisfies the equation.
But mind you, this is only the real solution to the equation.
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