Solving Radical Equations - Problem 3

This problem is kind of crazy-looking because not only is it 2 equal fractions, but it also involves square roots. For me whenever I see two equal fractions or a proportion, I like to use cross multiplying. Cross multiplying is where you write an equality statement as the product of the two diagonals. So I’ll have root 2 times 5 root 2, that’s these diagonals is going to be equal to the product of x times root 2 take away 1. Just be really careful with your parentheses, make sure x is being multiplied by both of those terms.

Okay well let’s go through and solve that. Root 2 time 5 root 2 is going to be 5 times 2 which is 10. 10 is equal to x times root 2 take away 1. Now usually I want to distribute something like this, but since x is being multiplied by some number like the square root of 2 take away 1 is like .4 or something, it’s some number and I’m going to want to get x all by itself. I’m not going to distribute. What I’m going to do at this step is divide both sides by the quantity root 2 take away 1 because now x is isolated. I know that x is equal to 10 over root 2 take away 1 and that’s kind of my answer, except that this is not proper form.

You guys know thou shall not have a radical on the denominator of a fraction, it’s just not simplified form. So the way I would simplify that answer is by multiplying by the Conjugate of the denominator. I’m going to multiply top and bottom by root 2 plus 1. The reason why that works is because when I FOIL on the bottom here, I’ll have just an integer on my denominator.

Here is what I mean. Root 2 times root 2 is regular old 2, that’s my firsts, outers I’ll have minus root 2, inners is plus root 2 so those guys are added inverses they become 0 at the end I need -1. On top of my fraction if I distribute the 10 I’ll have 10 root 2 plus 10. I’m almost done. The last thing I want to do is simplify that denominator there. This will be my final answer, 10 root 2 plus 10 divided by 1 which means I don’t have to write it. 10 root 2 plus 10 equals x, that’s my final answer.

Your textbook might have in the back of the book this 10 factored out, so it might say 10 parenthesis root 2 plus 1, it doesn’t matter those are equivalent statements.

So again guys cross multiplying is a great technique you learned months and months ago, but you can still use it to solve these problems. Be really careful when you get to something where x equals blah, blah, blah, make sure that blah, blah, blah is in reduced form. This guy needs to be multiplied by the conjugate in order to be simplified. I have to get that radical out of the denominator.