Question Video: Evaluating Algebraic Expressions Using the Difference of Two Squares Identity | Nagwa Question Video: Evaluating Algebraic Expressions Using the Difference of Two Squares Identity | Nagwa

# Question Video: Evaluating Algebraic Expressions Using the Difference of Two Squares Identity Mathematics

Given that π₯ = β(23) + β(7) and π₯π¦ = 1, find the value of π₯Β² β 256π¦Β².

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### Video Transcript

Given that π₯ equals the square root of 23 plus the square root of seven and π₯ times π¦ equals one, find the value of π₯ squared minus 256π¦ squared.

We know that π₯ equals the square root of 23 plus the square root of seven and that π₯ times π¦ equals one. If π₯ times π¦ equals one, then π¦ equals one over π₯. This means we can write π¦ as one over the square root of 23 plus the square root of seven. Generally, we wouldnβt want to write π¦ like this. We wouldnβt want to have the square roots in the denominator. To get the square root of 23 plus the square root of seven out of the denominator, we can rationalise. We do this by multiplying both sides of our equation by the square root of 23 minus the square root of seven. We do this by taking the square root of 23 minus the square root of seven over the square root of 23 minus the square root of seven and multiplying it by what we already had for π¦.

This fraction is equal to one because its numerator and its denominator are the same. This means weβre not changing the value of π¦. But why did we choose the square root of 23 minus the square root of seven? We do this because of the difference of squares property that tells us π squared minus π squared is equal to π plus π times π minus π. And so we can say that in our denominator we have π plus π times π minus π. And so when we start to simplify, in our numerator, we had one times the square root of 23 minus the square root of seven. So that stays the same.

But in our denominator, we can rewrite it as π squared minus π squared, which means we have the square root of 23 squared minus the square root of seven squared. The square root of 23 squared is 23. The square root of seven squared is seven. So in our denominator, we have 23 minus seven. 23 minus seven is 16. What weβre saying is that one over the square root of 23 plus the square root of seven is equal to the square root of 23 minus the square root of seven over 16. We now know our π₯ and π¦ values. And so weβre ready to solve for π₯ squared minus 256π¦ squared. If we look closely at π₯ squared minus 256π¦ squared, we can see that this expression is a difference of squares, where π equals π₯ and π equals 16π¦. 16 squared is 256 and π¦ squared is π¦ squared.

This means we can say that π₯ squared minus 256π¦ squared is equal to π₯ plus 16π¦ times π₯ minus 16π¦. And now, weβre ready to plug in what we know for π₯ and π¦. Our first variable is π₯. And we substitute the square root of 23 plus the square root of seven. And weβre going to add 16 times π¦, which is the square root of 23 minus the square root of seven over 16. For our second factor, everything is the same, except weβre subtracting 16π¦, instead of adding 16π¦. At this point we see that multiplying by 16 and a denominator of 16 cancels out on both sides. That means our first factor would then be the square root of 23 plus the square root of seven plus the square root of 23 minus the square root of seven.

In our second factor, we need to be careful because we are subtracting the square root of 23. And we are subtracting negative square root of seven, which will be adding the square root of seven. So we subtract the square root of 23 and add the square root of seven. What we can do now is check within the parentheses and see if thereβs anything that cancels out. Plus the square root of seven minus the square root of seven equals zero; they go away. Plus the square root of 23 minus the square root of 23 also goes away. The square root of 23 plus the square root of 23, we can write as two times the square root of 23.

The same thing goes for the square root of seven plus the square root of seven. We get two times the square root of seven. And then when we multiply them together. We can multiply two times two to get four. And how do we multiply the square root of 23 by the square root of seven? We multiply 23 by seven and then take the square root of that whole amount. 23 times seven equals 161. And so weβll have four times 161. π₯ squared minus 256π¦ squared under the given conditions is four times the square root of 161.

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