Video: Finding the Solution Set of Root Equations Using Factorisation

Find the solution set of √(5^(2𝑥 − 2) − (2/5) × 5^(𝑥) + 1) = 24.

04:12

Video Transcript

Find the solution set of the square root of five to the power of two 𝑥 minus two minus two-fifths times five to the power of 𝑥 plus one equals 24.

In finding the solution set to our equation, we’re looking to find the values of 𝑥, which make this equation true. And at first glance, this looks like a really horrible equation. But there’s a very simple thing that we can do to make it look a little nicer. And that is we’re going to square both sides of our equation by squaring the left-hand side, which is performing the inverse operation to square rooting. So, we’re just left with the values inside the square root. That is, five to the power of two 𝑥 minus two minus two-fifths times five to the power of 𝑥 plus one. Since we’re squaring the left side, we also need to square the right side. So that gives us 24 squared, which is 576.

Now, before dealing with these nasty exponents, let’s just begin by subtracting 576 from both sides. And so, our equation becomes five to the power of two 𝑥 minus two minus two-fifths times five to the power of 𝑥 minus 575 equals zero. And then, we’re going to look at our first term. And we’re going to use the laws of exponents that say that 𝑥 to the power of 𝑎 divided by 𝑥 to the power of 𝑏 can be written as 𝑥 to the power of 𝑎 minus 𝑏. And so, we’re going to use the inverse of this and say that “Well, five to the power of two 𝑥 minus two equals five to the power of two 𝑥 divided by five to the power of two.” And therefore, our equation is now five to the power of two 𝑥 over five squared minus two-fifths times five to the power of 𝑥 minus 575 equals zero.

Now, this does look a little bit like a quadratic equation. But before we deal with that bit, let’s get rid of these fractions. We’re going to multiply by the highest power of five in our denominators. Well, that’s five squared. Five to the power of two 𝑥 divided by five squared times five squared is just five to the power of two 𝑥. Two-fifths times five to the power of 𝑥 times five squared is two times five to the power of 𝑥 times five or 10 times five to the power of 𝑥. And 575 times 25 is 14375.

And we should now see this looks a little bit like a quadratic equation. We’re going to manipulate five to the power of two 𝑥 so it does. We’re going to write it as five to the power of 𝑥 squared. Remember, when we have an expression like this, we simply multiply the exponents. So, this is indeed the same as five to the power of two 𝑥. So, our equation is five to the power of 𝑥 squared minus 10 times five to the power of 𝑥 minus 14375 equals zero.

And now, we perform a substitution. We let 𝑦 be equal to five to the power of 𝑥. And so, our equation becomes 𝑦 squared minus 10𝑦 minus 14375 equals zero. And to solve this equation, we’re going to factor the expression on the left-hand side. Now, we know that the first term in each binomial must be 𝑦 since 𝑦 times 𝑦 gives us 𝑦 squared. And then, we look for two numbers whose product is negative 14375 and whose sum is 10. This will take a little bit of trial and error, but these values are negative 125 and 115. So, we see that 𝑦 minus 125 times 𝑦 plus 115 equals zero.

And for the product of these two expressions to be equal to zero, either one or the other of these expressions must themselves be equal to zero. So, either 𝑦 minus 125 equals zero or 𝑦 plus 115 equals zero. Let’s solve this first equation by adding 125 to both sides. And we find 𝑦 is equal to 125. Similarly, we solve our second equation by subtracting 115, and 𝑦 is equal to negative 115. We’re not quite finished though. We said we’re looking to find the values of 𝑥 which make our equation correct. So, we’re going to go back to our earlier substitution.

We replace 𝑦 with five to the power of 𝑥. And we see that five to the power of 𝑥 is either equal to 125 or negative 115. Well, we know that five cubed is 125. So, the solution to this first equation is 𝑥 equals three. However, there are no values of 𝑥 which will make the expression five to the power of 𝑥 negative. So, there are no solutions to our second equation. And that means our solution set consists of exactly one value. It’s three.