There isn't an "easy" factorisation for 5x^2+2x-1, so you can use the formula for the factors of
ax^2 + bx + b
(-b +- sqrt(b^2-4ac)) /2a
but I'm not quite sure why this proves the domain.
In the second part, I think I agree with you, but it depends where you put the brackets!
sqrt(y) - 10 has a domain y >= 0
sqrt(y-10) has a domain y-10 >= 0
Might that be the difficulty here?
The function 5x^2+2x-1's domain is all real numbers, according to the answer key. However, I don't know how to factor this to prove it. Could you show me?
The function the sq. root of y-10 has a domain of all numbers so that y is greater than or equal to zero. But I don't see how this is true. 1-10, for example would be -9, and this would be imaginary. Negative numbers would result in an imaginary value too, but I don't see how -1-10 is any more imaginary than 1-10.
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