How many roots does the following function have? Y=x^2 + 5

Question

anonymous · Answer

To know how many roots a second degree equation has, you use the formula: b^2-4ac. If it is negative, no roots, if it is 0, 1 root, and if it is >0 two roots. Where Y=ax^2+bx+c. Do you already learned this?

anonymous · Answer

Could I do it by doing a x y table and using -2,-1,0,1 and 2?

anonymous · Answer

That would be guessing, wich you cannot know it is going to work.

anonymous · Answer

Not exactly. This polynomial doesn't have real roots. How ivanmierner says you must to test the discriminant for this polynomial. A general form for a second degree of one variable polynomial is $$ax^{2}+bx+c=0$$ and the discrminant for that is $$b^{2}-4ac$$.

anonymous · Answer

You can see this by doing these steps:$$(ax^2+bx+c=0) 4a$$$$(4a^2x^2+4abx+4ac=0)+b^2$$$$4a^2x^2+4abx+b^2+4ac=(2ax+b)^2+4ac=b^2$$$$2ax+b=\pm \sqrt{b^2-4ac}$$If what is inside the square root is negative, x has no possible values, if it is 0, it has only 1 because +0=-0 and if it is positive, x has two values because there will be two values for the square root.

anonymous · Answer

WOW! the posts changed order...

jhonyy9 · Answer

so

y=x^2 +5 for you can calcule the roots of this function make it equal with zero

x^2 +5=0 so than 
x^2 = -5 than 

x_1,2=+/- sqrt(-5) = +/- isqrt5 so result that has roots just complex numbers 

ok ?