# Video: Using Binomial Expansion to Solve Exponential Equations

Consider the expansion of ((𝑎/𝑥⁴) + 𝑥)¹⁰. Given that the constant of this expansion is 720, find all the possible values of 𝑎.

04:14

### Video Transcript

Consider the expansion of 𝑎 over 𝑥 to the fourth power plus 𝑥 to the 10th power. Given that the constant of this expansion is 720, find all possible values of 𝑎.

Here, we have a binomial raised to a positive integer power. We’re given some information about the constant value in its expansion. So, that tells us we’re going to need to consider the binomial expansion of 𝑎 over 𝑥 to the fourth power plus 𝑥 to the 10th power. Let’s consider a binomial 𝑏 plus 𝑐. Now, you might also see this written as 𝑎 plus 𝑏 or 𝑥 plus 𝑦. I’ve chosen 𝑏 plus 𝑐 so it doesn’t interfere with the values in our question.

The expansion of 𝑏 plus 𝑐 to the 𝑛th power is the sum from 𝑘 equals zero to 𝑛 of 𝑛 choose 𝑘 times 𝑏 to the power of 𝑛 minus 𝑘 times 𝑐 to the 𝑘th power. Now, this can be a little confusing to work with, so you might also consider its expanded form. We start with 𝑏 to the 𝑛th power. Our next term is 𝑛 choose one 𝑏 to the power of 𝑛 minus one times 𝑐. Our next term is 𝑛 choose two 𝑏 to the power of 𝑛 minus two 𝑐 squared. And we continue that pattern all the way up to 𝑐 to the 𝑛th power.

Notice we reduce the power of 𝑏 by one each time and increase the power of 𝑐. Now, whilst we’re not going to consider the full binomial expansion, we will consider quite a few pieces. So, let’s define 𝑏, 𝑐, and 𝑛. We see that 𝑏 is 𝑎 over 𝑥 to the fourth power. We could define 𝑐 as 𝑥. And 𝑛 is equal to 10. Remember, for this binomial theorem to work, 𝑛 must be a positive integer. And so, the first term in our expansion is 𝑏 to the power of 𝑛. So, that’s 𝑎 over 𝑥 to the fourth power to the power of 10.

Remember, we’re looking for a constant term. That’s a term without any 𝑥’s in it. We can see quite clearly that if we distribute the 10 over our fraction, we’re still going to have an 𝑥. So, what about our second term? Well, we have 10 choose one times 𝑎 over 𝑥 to the fourth power to the ninth power times 𝑥. Once again, if we distribute the nine over 𝑎 over 𝑥 to the fourth power, we end up with an even higher power of 𝑥 on our denominator. There’s no way that 𝑎 to the ninth power over 𝑥 to the power of 36 times 𝑥 will give us a constant. So, we’re actually gonna go to the other end of our expansion.

We know that the final term will be 𝑐 to the 𝑛th power. So, that’s 𝑥 to the 10th power, still not a constant. The term before this will be 10 choose nine times 𝑎 over 𝑥 to the fourth power times 𝑥 to the ninth power. Even if we cancel by 𝑥 to the fourth power, we end up with a coefficient of 𝑥 to the fifth power, so still no constants. The term before this, however, is 10 choose eight times 𝑎 over 𝑥 to the fourth power squared times 𝑥 to the power of eight. When we distribute this two over 𝑎 over 𝑥 to the fourth power, we get 𝑎 squared over 𝑥 to the eighth power. Remember, we multiply these exponents.

We then see that 𝑥 to the eighth power divided by 𝑥 to the eighth power is one. So, this term is simply 10 choose eight times 𝑎 squared; it’s our constant term. And according to the question, that’s equal to 720. So, we can say that 10 choose eight times 𝑎 squared must be equal to 720. Let’s evaluate 10 choose eight.

𝑛 choose 𝑟 is 𝑛 factorial over 𝑟 factorial times 𝑛 minus 𝑟 factorial. And this means that 10 choose eight is 10 factorial over eight factorial times 10 minus eight, which is two factorial. Of course, we know we can rewrite 10 factorial as 10 times nine times eight times seven and so on, or 10 times nine times eight factorial. Then, we divide through by eight factorial. Two factorial is simply two times one, so we can divide through by two. And we’re, therefore, left with 10 choose eight as being equal to five times nine all over one, which is simply 45.

And so, our equation becomes 45𝑎 squared equals 720. Next, we divide through by 45. And 720 divided by 45 is 16. So, we find 𝑎 squared is equal to 16. The last thing that we need to do to solve for 𝑎 is to find the square root of both sides of our equation, remembering to find both the positive and negative square root of 16. The square root of 16 is four. And so, 𝑎 is either positive or negative four.