SOLUTION: How can I solve 2 square root of 7t times square root of 14t? I came out with 2 square root of 98t squared, then 2 square root of 49 times 2t squared followed by 2 times 7 square

Algebra ->  Radicals -> SOLUTION: How can I solve 2 square root of 7t times square root of 14t? I came out with 2 square root of 98t squared, then 2 square root of 49 times 2t squared followed by 2 times 7 square       Log On


   

Question 73436: How can I solve 2 square root of 7t times square root of 14t? I came out with 2 square root of 98t squared, then 2 square root of 49 times 2t squared followed by 2 times 7 square root of 2. My answer was 14t square root of 2. Is this correct? I don't know how to write out the problem, I hope this is okay.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
2%2Asqrt%287t%29%2Asqrt%2814t%29
.
Simplify this.
.
Let's see if your answer is correct.
.
One way to do this is to use the rule: sqrt%28a%2Ab%29+=+sqrt%28a%29%2Asqrt%28b%29
.
Applying this rule to 7t leads to sqrt%287t%29+=+sqrt%287%29%2Asqrt%28t%29
.
Then apply this rule twice to 14t to get:
.
sqrt%2814t%29+=+sqrt%282%2A7%29%2A+sqrt%28t%29+=+sqrt%282%29%2Asqrt%287%29%2Asqrt%28t%29
.
Now substitute for sqrt(7t) and sqrt(14t) in the original problem to get:
.
2%2Asqrt%287%29%2Asqrt%28t%29%2Asqrt%282%29%2Asqrt%287%29%2Asqrt%28t%29
.
In this string we have sqrt%287%29%2Asqrt%287%29 and that equals 7. We also have
sqrt%28t%29%2Asqrt%28t%29 and that equals t. The rest of the terms remain unchanged.
.
So, after substituting t and 7 the problem reduces to:
.
2%2A7%2At%2Asqrt%282%29
.
which further simplifies to:
.
14%2At%2Asqrt%282%29
.
You might want to change it to have the variable last. If you do the answer is:
.
%2814%2Asqrt%282%29%29%2At
.
But there's nothing wrong with your answer. Another good job!