SOLUTION: square root of 11(square root of 11 + x square root 22) How do i do this kind of problem? I am lost it is square root of 11 times square root of 11 plus x square root of

Algebra ->  Square-cubic-other-roots -> SOLUTION: square root of 11(square root of 11 + x square root 22) How do i do this kind of problem? I am lost it is square root of 11 times square root of 11 plus x square root of       Log On


   



Question 98208: square root of 11(square root of 11 + x square root 22)
How do i do this kind of problem? I am lost it is
square root of 11 times square root of 11 plus x square root of 22

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
sqrt%2811%29%2A%28sqrt%2811%29+%2B+x%2Asqrt%2822%29%29
.
This is a problem involving distributed multiplication. It is of the form:
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A%2A%28B+%2B+C%29+=+A%2AB+%2B+A%2AC <=== call this rule 1
.
Notice that you are just multiplying the term outside the parentheses by each of the terms
inside the parentheses.
.
What makes this problem a little "different" is that it involves multiplication of square roots,
so you need to know a few rules about those multiplications. Here are a couple of rules for
square roots:
.
sqrt%28A%29%2A+sqrt%28A%29+=+A <=== call this rule 2
.
and:
.
sqrt%28A%2AB%29+=+sqrt%28A%29%2Asqrt%28B%29 <=== call this rule 3
.
It seems as if this problem involves a lot of sqrt%2811%29 terms. But notice that there
is one term that is sqrt%2822%29. If we apply rule 3 to that term, we can get a sqrt%2811%29
from it by doing the following:
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sqrt%2822%29+=+sqrt%282%2A11%29+=+sqrt%282%29%2Asqrt%2811%29
.
So in the problem we can replace sqrt%2822%29 by sqrt%282%29%2Asqrt%2811%29 and the problem
then becomes:
.
sqrt%2811%29%2A%28sqrt%2811%29%2B+x%2Asqrt%282%29%2Asqrt%2811%29%29
.
Now let's apply rule 1. We'll do that by multiplying sqrt%2811%29 times each of the
terms in the parentheses to get:
.
sqrt%2811%29%2Asqrt%2811%29+%2B+sqrt%2811%29%2Ax%2Asqrt%282%29%2Asqrt%2811%29
.
Now we can apply rule 2 to both of these terms. The first term is sqrt%2811%29%2Asqrt%2811%29
and rule 2 tells us that the answer is just 11. So substituting 11 into the problem reduces
the problem to:
.
11+%2B+sqrt%2811%29%2Ax%2Asqrt%282%29%2Asqrt%2811%29
.
Notice now that in the second term there is also a sqrt%2811%29%2Asqrt%2811%29 involved. Rule
2 also allows us to replace that product with 11. And this substitution further reduces the
problem to:
.
11+%2B+11%2Ax%2Asqrt%282%29
.
That's the answer to this problem. I suppose that if you wanted to you could factor out
the 11 that is common to both terms to get 11%2A%281+%2B+x%2Asqrt%282%29%29 but that's up to you
and what your instructions are.
.
Hope this helps you to understand the problem a little more.