The triangles are similar, solve for K.
2 answers:
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The equation given is 
where a is some number.
We can solve for x by taking the square root of both sides.
Now let's think through what happens for various values of a.
TWO SOLUTIONS
If a is a positive number the above yields two solutions. Take for example:
There will be two solutions (one positive and one negative) as there are two numbers (here -7 and +7) that when multiplied by themselves give 49. That is,
and 
. The positive root is called the principal root and the negative root is called the secondary root. This will be the case anytime we take the root of a positive number.
ONE SOLUTION
If a = 0 there is only one solution. That is because
and 
. Zero is neither positive nor negative and it has only one root which is 0 itself. So in this case there is only one solution and it is 0.
NO (REAL) SOLUTIONS
If a is negative we would be taking the square root of a negative number. There is no (real) number that when multiplied by itself gives a negative number. Take for example
which gives us 
. The square root of -49 is not 7 because (7)(7)=49 which is positive. The square root of -49 is not -7 because (-7)(-7)=49 which is also positive. There is no real number that gives -49 when multiplied by itself. I say "real" numbers because there do exist imaginary/complex numbers but because of the way the questions was asked I imagine you may not know about these yet.
We can solve for x by taking the square root of both sides.
Now let's think through what happens for various values of a.
TWO SOLUTIONS
If a is a positive number the above yields two solutions. Take for example:
There will be two solutions (one positive and one negative) as there are two numbers (here -7 and +7) that when multiplied by themselves give 49. That is,
ONE SOLUTION
If a = 0 there is only one solution. That is because
NO (REAL) SOLUTIONS
If a is negative we would be taking the square root of a negative number. There is no (real) number that when multiplied by itself gives a negative number. Take for example
First of all, we can observe that
So the expression becomes
This means that the expression is defined for every
Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask
Since we can't accept 3 as an answer, the actual solution set is